1. No category

Momentum
Momentum is defined as a quantity of motion equal to the product of the mass and the velocity of
an object. The following equation is the definition of momentum stated mathematically:


p  mv


where p is the momentum of the object (kg.m/s), m is the mass of the object (kg), and v is the
velocity of the object (m/s). Because mass is a scalar and velocity is a vector, the product of
mass and velocity (momentum) is a vector and is in the same direction as the velocity.
A vector quantity is a quantity that has a direction associated with it. It is important for a
direction to be included with momentum. Momentum is therefore a vector quantity.
Example 9.1 p.451
Practice Problems p.451
Example 9.2 p.452
Practice Problems p.452
Another Example Not in Your Book:
A 1.50 kg object falls from a height of 14.0 m. What is the momentum of the object as it hits the
ground?
Check and Reflect p.453 #1-2, 5-9,11,13-16
MC #1-2
B. Dickie
1
Impulse
You have learned Newton’s second law in the following form:


F  ma


where F is the net force on an object (N), m is the mass of the object (kg), and a is the
acceleration of the object (m/s2).
Through some substitutions and rearranging:


F  ma


v
F m
t


Ft  mv

So p , which represents the change in momentum, is equal to the net force on an object
multiplied by the time that the force is applied, and the mass of the object multiplied by the
change in velocity of the object. This change in momentum quantity can also be referred to as

impulse, which can be known by the symbol J . Impulse has the same units as momentum.
Example 9.3 p.458
Practice Problems p.458
Force vs. Time graphs are sometimes used to describe impulse imparted on an object. To find
the impulse imparted on an object from a F vs. t graph, simply calculate the area between the line
and the x-axis. If the area is above the x-axis, consider the impulse to be positive (momentum
gained by the object). If the area is below the x-axis, consider the impulse to be negative
(momentum lost by the object). The impulse found by calculating the area can be set equal to
either part of the impulse equation to solve for another variable.
Example 9.4 p.462
Practice Problems p.462
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This impulse equation implies that a force in contact with an object for a greater period of time
will result in a greater change in momentum (impulse) and therefore a greater velocity. A good
golf swing or a good baseball swing does not need to impart a greater force on the ball than a
poor swing, it simply requires the club or the bat to be in contact with the ball for a greater
period of time.
The impulse equation also has implications for objects that have a predetermined change in
momentum (impulse). A person, for example, jumping from a ledge must come to a stop when
he / she hits the ground. The person’s mass and change in velocity are predetermined, resulting
in a predetermined change in momentum (impulse). If the predetermined impulse is spread out
over a greater time interval, however, the force acting on the person’s legs will be less than it
would be if the impulse were spread out over a shorter time interval. To spread the impulse over
a greater time interval, one needs to simply bend his / her legs upon landing. The next time you
jump off of something, take note that you intuitively bend your legs when you land in order to
prevent injury. An airbag in a car is additional application of predetermined impulse. When you
are in a car accident, the mass and change in velocity of the car are predetermined. The impulse
is therefore predetermined as well. If, however, it is possible to spread the impulse out over a
longer time interval, the force acting on the driver’s body will decrease resulting in less
probability of injury. An airbag is designed to do just that, spread the predetermined impulse
over a longer period of time. Any safety device designed to “cushion a blow” is really designed
to spread a predetermined impulse out over a longer period of time, resulting in a lesser force.
Check and Reflect p.453 #3-4
Check and Reflect p.467 #1-13
MC #3-9
B. Dickie
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The Law of Conservation of Momentum
The law of conservation of momentum states that the total momentum of all objects in an
isolated system always remains constant. In other words, if one can isolate a system from
friction, wind, etc. then the total momentum within that system will never change. Some
implications of this law are as follows:
-the sum of the momenta of all colliding objects before a collision will be the
same as the sum of the momenta after the collision
-the total momentum before an explosion (usually 0 kg.m/s unless the exploding
object is moving before the explosion) is the same as the total momentum of all
explosion fragments after the explosion
The law of conservation of momentum is a vector conservation law because one needs to
consider and account for the directions of all of the colliding objects before and after a collision.
The concept of the law of conservation of momentum is valid for a given problem if one of the
following criteria is met:
1. two or more objects collide with one another
2. there is an explosion of some sort where one object breaks apart into two or more objects.
There are two ways in which the law of conservation of momentum can be used: onedimensional interactions (collisions or explosions) and two-dimensional interactions (collisions
or explosions). Remember that direction is important for both, and that it must be taken into
account every time the law of conservation of momentum is used.
The following are some suggested sign conventions for accounting for directions in conservation
of momentum problems:
1. The velocity of objects moving North, East, up, or right should be considered to be positive
(+).
2. The velocity of objects moving South, West, down, or left should be considered to be
negative (-).
Problem solving strategy for one-dimensional problems:
1.
2.
3.
4.
Write down the givens and the variable to be solved for.


Write down “ pi  p f ”.
Substitute “mv” in for each mass for before and after the interaction.
Plug in the numbers and solve.
B. Dickie
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Example 9.8 p.479
Practice Problems
Example 9.6 p.477
Practice Problems
Example 9.5 p.476
Practice Problems
Example 9.7 p. 478
Practice Problems
MC #10-14
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Elastic and Inelastic Collisions
An elastic collision is one in which kinetic energy is conserved (i.e. the total kinetic energy of all
of the colliding objects after the collision is the same as the total kinetic energy of all of the
colliding objects before the collision). This is a scalar conservation law, because the directions
of all of the colliding objects before and after a collision are not important.
An inelastic collision is one in which kinetic energy is lost (i.e. the total kinetic energy of all of
the colliding object after the collision is less than the total kinetic energy of all of the colliding
objects before the collision). Note that momentum is still conserved. Total momentum is never
lost in a collision as long as the system is isolated.
In the real world, no collision is truly elastic. There is always some kinetic energy lost in any
collision. There are, however, examples of collisions in the real world that closely approximate
elastic collisions. The collision of two billiard balls is an example of a nearly elastic collision.
The collision of two automobiles is an example of an inelastic collision in which a significant
amount of kinetic energy is typically lost. The energy lost would typically be mostly in the form
of heat.
To determine whether a collision is elastic or inelastic, calculate the total kinetic energy of all of
the objects before the collision. Remember that since kinetic energy is a scalar, no negative
signs or directions or vector components are needed. Then calculate the total kinetic energy of
all of the objects after the collision. Again, remember that since kinetic energy is a scalar, no
negative signs or directions or vector components are needed.
Example 9.9 p.482
Practice Problems
Example 9.10 p.484
Practice Problems
Example 9.11 p.485
Practice Problems
Check and Reflect p.486 #1-11
MC #15-16
B. Dickie
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More on the Law of Conservation of Momentum
When analyzing a two-dimensional interaction, one must consider both the x-components and ycomponents of all velocities both before and after the collision or explosion. The x- and ycomponents of the velocities can be found using some trigonometry The law of conservation of
momentum holds for both components and the same sign conventions can be used for both as are
used in one-dimensional interactions. In effect, breaking a two-dimensional interaction up into
x- and y-components simplifies the problem into two separate one-dimensional interactions.
The results of the x-interaction and the y-interaction are simply combined using a vector diagram
and mathematics (Pythagorean Theorem and the inverse tangent function).
Example 9.13 p.492
Practice Problems
Example 9.12 p.490
Practice Problems
Example 9.14 p.494
Practice Problems
Example 9.15 p.496
Practice Problems
Check and Reflect p.499 #1-3, 5-11
MC #17-19
Unit Review MC #20-31
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Diploma Exam Review Questions
Momentum
Use the following information to answer the next question.
1. What is the magnitude of the momentum of the baseball the instant before it reaches the ground?
A. 1.29 kg•m/s
B. 1.41 kg•m/s
C. 2.60 kg•m/s
D. 2.90 kg•m/s
-----------------------------------------------------------------Use the following information to answer the next question.
2. The momentum of the supertanker at cruising speed, expressed in scientific notation, is b 10w kg.m/s. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
------------------------------------------------------------------
Impulse
3. In an automobile collision, the severity of injury to the driver can be reduced by an airbag. In a car initially traveling at 100 km/h, the airbag
stops a 62 kg driver in 90 ms. The magnitude of average force exerted by the airbag on the driver is
A. 6.9 x 104 N
B. 1.9 x 104 N
C. 9.6 x 103 N
D. 6.1 x 102 N
4. The SI units for impulse may be written as
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Use the following information to answer the next question.
5. Assuming the maximum speed of a 20.6 g arrow, when released from full draw, is 94.0 m/s, the magnitude of the impulse that the bow gives
to the arrow when released from full draw is __________ N•s.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
------------------------------------------------------------------
Use the following information to answer the next question.
6. A satellite has a mass of 172 kg. To correct its orbit, a thruster is fired for 2.27 s, which changes the velocity of the satellite by 5.86 10–3 m/s.
The force generated by the thrusters, expressed in scientific notation, is b 10–w N. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
------------------------------------------------------------------
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Use the following information to answer the next two questions.
7. The magnitude of the impulse delivered by the hammer to the pile is
A. 61.8 kN•s
B. 30.9 kN•s
C. 7.46 kN•s
D. 3.73 kN•s
8. The impulse is delivered by this pile-driver in 2.10 10–3 s. The magnitude of the force that the hammer exerts on the pile, expressed in
scientific notation, is b 10w N. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
------------------------------------------------------------------
9. A rock climber falls and is saved from injuries by a climbing rope that is slightly elastic. The importance of the elasticity of the climbing rope
can be understood in
terms of impulse because elasticity results in
A. decreased force during an increased time interval
B. increased force during an increased time interval
C. decreased force during a decreased time interval
D. increased force during a decreased time interval
Conservation of Momentum (1D)
10. A 2 100 kg van collides with a 1 200 kg car that is at rest. They lock together and move together at a speed of 4.50 m/s. The initial speed of
the van is __________ m/s.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
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Use the following information to answer the next question.
11. Assume that the white ball is moving at a speed of 3.13 m/s when it collides with the black ball, which was initially at rest. The white ball
continues in its original direction. The speed of the white ball after the collision is 0.147 m/s. The speed of the black ball immediately after the
collision is __________ m/s.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
------------------------------------------------------------------
Use the following information to answer the next question.
12. The two lumps stick together, and no external horizontal forces act on the system. The velocity of the combined lump after the collision is
A. 60.0 cm/s, south
B. 31.7 cm/s, south
C. 20.0 cm/s, north
D. 15.0 cm/s, north
------------------------------------------------------------------
13. An empty freight car of mass m coasts along a track at 2.00 m/s until it couples to a stationary freight car of mass 2m. The final speed of the
two freight cars immediately after collision is
A. 1.50 m/s
B. 1.33 m/s
C. 1.15 m/s
D. 0.667 m/s
14. Two boys, Ted and Larry, initially at rest, push each other apart on a frictionless surface. Ted has a mass of 40 kg and Larry has a mass of 60
kg. After the boys push each other apart, Ted has a speed of 6 m/s. As the boys move apart, Larry has
A. more momentum than Ted
B. less momentum than Ted
C. more kinetic energy than Ted
D. less kinetic energy than Ted
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Elastic / Inelastic Collisions
15. The following statements all relate to a collision between any two objects on a horizontal frictionless surface. Which of these statements is
always true?
A. The kinetic energy of each object before and after the collision is the same.
B. The momentum of each object before and after the collision is the same.
C. The total momentum of the two objects before and after the collision is the same.
D. With respect to the surface, the gravitational potential energy of each object before and after the collision increases.
Use the following information to answer the next question.
16. Which of the following statements best describes the inelastic collision shown above?
A. Momentum is not conserved, and kinetic energy is not conserved.
B. Momentum is conserved, but kinetic energy is not conserved.
C. Momentum is not conserved, but kinetic energy is conserved.
D. Momentum is conserved, and kinetic energy is conserved.
------------------------------------------------------------------
Conservation of Momentum (2D)
Use the following information to answer the next two questions.
B. Dickie
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17. The magnitude of the momentum of the third piece of glass, fragment C, is
A. 5.19 kg•m/s
B. 3.85 kg•m/s
C. 2.28 kg•m/s
D. 0.610 kg•m/s
18. The speed of the third fragment of glass, expressed in scientific notation, is b _ 10 w m/s. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
-----------------------------------------------------------------Use the following information to answer the next question.
19. The speed of the red rock, after contact, is
A. 0.15 m/s
B. 0.22 m/s
C. 0.33 m/s
D. 0.39 m/s
Unit 1 Mixed Review Questions
20. Which of the following quantities is a scalar quantity?
A. Force
B. Power
C. Impulse
D. Momentum
21. A golf ball has a mass of 45.0 g. A golf club is in contact with the golf ball for 3.00 × 10 –4 s, and the ball leaves the club with a speed of 72.0
m/s. The average force exerted by the club on the ball, expressed in scientific notation, is b × 10w N. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
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22. In a vehicle safety test, a 1 580 kg truck traveling at 60.0 km/h collides with concrete barrier and comes to a complete stop in 0.120 s. The
magnitude of the change in the momentum of the truck, expressed in scientific notation, is b × 10w kg•m/s. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
Use the following information to answer the next question.
23. The speed of the hopper car immediately after receiving the load of coal, expressed in scientific notation, is b x 10 -w m/s. The value of b is
__________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
------------------------------------------------------------------
Use the following information to answer the next question.
24. The velocity of the 2.4 kg object after collision is
A. 15 m/s to the right
B. 8.7 m/s to the left
C. 8.0 m/s to the right
D. 6.2 m/s to the left
25. Two carts, each with a spring bumper, collide head-on. At one point during the collision, both carts are at rest for an instant. At that instant,
the kinetic energy that the carts originally possessed is almost completely
A. lost to friction
B. transformed into heat and sound
C. converted into kinetic energy in the spring bumpers
D. converted into potential energy in the spring bumpers
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26. A 1 575 kg car, initially traveling at 10.0 m/s, collides with a stationary 2 250 kg car. The bumpers of the two cars become locked together.
The speed of the combined cars immediately after impact is __________m/s.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
27. A 115 g arrow traveling east at 20 m/s imbeds itself in a 57 g tennis ball moving north at 42 m/s. The direction of the ball-and-arrow
combination after impact is
A. 4N of E
B. 4E of N
C. 2E of N
D. 2N of E
28. In an inelastic collision, the energy that appears to be missing is converted into
A. sound and momentum
B. force and momentum
C. sound and heat
D. heat and force
29. Which of the following units are correct units for momentum?
A. J•s
B. N•m
C. N•s
D. N/J
Use the following information to answer the next two questions.
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30. The magnitude of the impulse on the car during the test drive is
A. 4.80 × 103 kg•m/s
B. 1.92 × 104 kg•m/s
C. 2.40 × 104 kg•m/s
D. 2.88 × 104 kg•m/s
Use your recorded answer from the previous question to answer the next question.
31. The average net force on the car during the test drive, expressed in scientific notation, is a.bc × 10d N. The values of a, b, c, and d are ____,
____, ____, and ____.
(Record all four digits of your answer in the numerical-response section on the answer sheet.)
------------------------------------------------------------------
Momentum MC Answers
1
2
3
4
5
D
5.56
B
B
1.94
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7
8
9
10
4.44
C
3.55
A
7.07
11
12
13
14
15
3.27
D
D
D
C
16
17
18
19
20
B
C
1.17
C
B
21
22
23
24
25
1.08
2.63
9.09
B
D
26
27
28
29
30
31
4.12
A
C
C
B
1503
16
Electric Charge
There are two types of electric charge: positive and negative. Positive charges are protons,
which are located in the nucleus of the atom. Negative charges are electrons, which orbit the
nucleus of the atom. The unit of electric charge is the Coulomb ( C ). An electron has a charge
of -1.60 x 10-19 C.
Some materials have electrons that are tightly bound to the nucleus and are not free to travel
within the substance. These materials are called insulators. Conductors, on the other hand, are
materials that have electrons in the outermost regions of the atom that are free to travel.
(Pearson Physics, Ackroyd et al, 2009)
The Laws of Electric Charge
Like charges repel (push away) while unlike charges attract (come together). Charged objects
also attract neutral objects, but this is not one of the two fundamental laws; rather it is a result of
the two fundamental laws. Charged objects attract neutral objects because of an induced
temporary separation of charge (discussed in detail later).
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The Law of Conservation of Charge
The law of conservation of electric charge states that the total amount of charge in a closed
system is constant. This means that electric charge cannot be created or destroyed. Negative
charges can, however, be transferred.
If two neutral objects are rubbed together, negative charges can be transferred from one to the
other. The object that gives up the negative charges becomes positively charged, while the
object that accepts the negative charges becomes negatively charged. The magnitudes of the net
charges will be equal, although one will be positive and one will be negative.
If two objects with unequal charges are touched, negative charges can be transferred from the
object that has the higher negative charge (or lower positive charge) to the object that has the
lower negative charge (or higher positive charge). The “goal” is for the charges of both objects
to balance out. For example, a ball with a charge of –2.0 C is touched with a ball with a charge
of +4.0 C. Negative charges are transferred from the first ball to the second, so that the final net
charge of both balls is +1.0 C.
Remember that it is always the negative charges that are transferred, not the positive.
An Example Not in Your Book
A ball (ball #1) with a charge of +3.0 C touches a ball (ball #2) with a charge of –1.0 C. The
first ball then touches a third ball (ball #3) with a charge of –5.0 C. What is the final charge on
all three balls?
MC #32-33
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Charging Objects
There are two methods of transferring charge: conduction and induction. Conduction refers to a
charge transfer resulting from two objects touching, while induction refers to charging resulting
from two objects being brought close to one another, but not touching.
Charging by conduction can occur in one of two ways: friction or contact.
Charging by friction occurs when two charged or neutral objects are rubbed together. Negative
charges transfer from one object to the other, resulting in one of the objects becoming more
negative (or less positive) and one becoming less negative (or more positive). When charging by
friction, the initial charges of each object are not important. To determine which of the two
materials loses the electrons and which one gains, we use the electrostatic series. It tells us
which materials hold electrons tightly and which ones hold electrons loosely.
(Pearson Physics, Ackroyd et al, 2009)
Charging by contact (often simply called charging by conduction) occurs when two objects
touch (or come close enough to touching that electrons jump), with at least one of the objects
being initially charged. The object that has more negative charge (or less positive charge)
transfers negative charge to the other until the charges on both objects become balanced
(although this does depend somewhat on the sizes and shapes of the objects). When charging by
contact, the initial charges of each object must be different.
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Charging by induction can result in a temporary separation of charge or a permanent charge on
an object.
When a charged object is brought near a neutral object, negative charges both objects
redistribute. The negative charges go as far away from other negative charges and as close to
positive charges as they can. This results in a temporary separation of charge. This is the
reason why charged objects are attracted to neutral objects.
Once the charged object is removed, the negative charges redistribute throughout the formerly
neutral object.
To permanently charge an object by induction, a ground wire must be introduced. This gives the
object access to an essentially infinite source / drain of electrons (literally the ground!). Always
remember that it is only negative charges that can move.
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An electroscope can be used to detect charge. An electroscope typically consists of an insulated
see-through container with a metal shaft inserted into it, and a metal plate or metal ball attached
to the top of the shaft. The bottom of the shaft has two thin pieces of foil attached to it. The
same principles apply to charging an electroscope by induction as have been previously
described.
An electroscope will detect both positive and negative charge, but it cannot, by itself, distinguish
between them.
Check and Reflect p.523 #5-12
MC #34-39
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Coulomb’s Law
Coulomb was aware that there was a force between two charged particles but was unaware of
how to calculate it. He devised a clever experiment called a torsion balance experiment to
determine an equation to find the strength of the electric force. Coulomb’s torsion balance
experiment is similar to the later developed experiment to determine the gravitational constant.
That experiment is known as Cavendish’s torsion balance experiment.
Coulomb brought a charged ball near a charged object on the end of the arm of the torsion
balance. The resulting electrostatic force caused the arm to rotate, causing a sensitive spring to
tighten or loosen, ultimately causing a needle to move a proportionate angle. The amount of
movement of the needle was related to the electrostatic force between the two charged objects.
Coulomb had no way of measuring the charges involved in his experiment, so he devised a
clever way to determine the relative charges. He started with an object of arbitrary charge q. He
knew that when he touched this with an uncharged object, they would both have a charge of ½q.
Manipulating and measuring the distance between the charges, and measuring the resulting force
each time, he was able to determine that the force was inversely related to the square of the
separation distance:
F
1
r2
He then held the separation distance constant and manipulated the charge by touching objects
(and measuring the resulting relative charges), and measured the resulting electrostatic force.
From this, Coulomb was able to determine that the force was directly related to the product of
the charges:
Fq1q2
Therefore,
F
q1 q 2
r2
Adding a proportionality constant to make the above an equation, we get:
F k
q1 q 2
r2
where F is the electric force (N), k is a constant known as Coulomb’s constant (8.99 x 109
N.m2/C2), q1 and q2 are the magnitudes of the charges experiencing each other’s force (C), and r
is the distance between the two charges (m).
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Note that you should not substitute the sign of the charge into the above equation. Even for
negative charges, you should substitute a positive value. Also note that electric force is a vector
quantity.
Coulomb’s
law
Newton’s
gravitation
relationship
inverse square
magnitude
large (k~109)
inverse square
small (G~10-11)
direction
attractive /
repulsive
attractive
Example 10.1 p.530
Practice Problems p.530
Example 10.2 p.531
Practice Problems p.531
MC #40-43
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More on Coulomb’s Law
If there are three or more charges interacting with each other, calculating the force on one of the
charges caused by the other two can be done with Coulomb’s law, albeit in a slightly more
complicated way than described above.
Calculating the electrostatic force when there are three or more charges requires the use of
vectors. If all of the charges are in a line, the following steps should be used to solve for the
force:
1. Draw a free body diagram of the object for which you are attempting to calculate the net
force.
2. Using Coulomb’s law, calculate the individual forces acting on the object.
3. Add the individual forces together, making sure you take into account the directions of the
forces (with positive and negative signs).
If the charges are not in a line, the following steps should be used to solve for the force:
1. Draw a free body diagram of the object for which you are attempting to calculate the net
force.
2. Using Coulomb’s law, calculate the individual forces acting on the object.
3. Draw a vector diagram using all of the forces above. Use the Pythagorean Theorem to solve
for the magnitude of the force and the inverse tangent function to solve for the direction of
the force.
Example 10.3 p.532
Practice Problems p.532
Example 10.4 p.533
Practice Problems p.533
Example 10.5 p.534
Practice Problems p.534
Example 10.6 p.535
Practice Problems p.536
Then, Now, and Future: ESD Control Manager p.537 #1-3
Check and Reflect p.538 #1-10
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Electric Fields
A field is a line of influence surrounding something. There are two types of fields: scalar and
vector fields. A scalar field is a line of influence around something that has no direction
associated with it. For example, the light coming from a light bulb influences a person, but not
in any particular direction. Light is therefore a scalar field. The same is true of the heat that
comes from a space heater. A vector field, on the other hand, is a line of influence around
something that has a direction associated with it. For example, the Earth has a gravitational field
surrounding it. The Earth’s gravitational field attracts objects toward it. The direction of the
influence is important here, so gravitational fields are therefore vector fields. A similar argument
holds for electric fields, attractive or repulsive lines of influence surrounding a charged particle.
The direction of the electric field is defined as the direction in which a positive test charge would
move if placed in that field. In other words, electric fields always point away from positive and
toward negative.
A force only exists if there is something to experience it. A field is always present, whether
anything experiences it or not. Think of the Earth’s gravitational field. It is there whether or not
it is pulling on anyone or anything. The same is true for electric fields. Forces are measured in
Newtons, whereas fields are measured in Newtons per unit mass or charge, depending on the
type of field.
Below are some diagrams of the field lines surrounding various systems of charges:
a.
b.
POS
NEG
c.
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d.
e.
f.
pos
neg
Note that in every case, field lines point away from positive charge and toward negative charge.
Also note that all of the above fields are non-uniform except for the parallel plate diagram (f).
The electric field between two parallel plates is uniform. Electric fields are uniform if the field
lines are parallel to each other.
MC #44-45
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More on Electric Fields
The strength of an electric field at a point r metres from the charged object that produced it can
be found by the following equation:

q
E  k 12
r

where E is the electric field strength (N/C), k is Coulomb’s constant (8.99 x 109 N.m2/C2), q1 is
the charge of the object producing the field (C), and r is the distance from the field for which one
wishes to find the strength of the field (m).
Note that there does not have to be anything experiencing this field in order to use this equation.
The strength of an electric field experienced by a charged object can be found using the
following equation:

 Fe
E
q


where E is the electric field experienced by a charged object (N/C), Fe is the electric force
experienced by a charged object (N), and q is the magnitude of the charge experiencing the field
(C).
Note that you should not substitute the sign of the charge into the above equation. Even for
negative charges, you should substitute a positive value.
One must consider the basic definition of the direction of electric field in order to find its
direction. The direction of electric field is defined as “the direction in which a positive test
particle would move if placed in that field.” In other words, if the producer is positive, the field
lines point away from it. If the producer is negative, the field lines point towards it.
If one wishes to find the net electric field strength at a point due to two or more producers, a
diagram is usually an appropriate way to start. The diagram should show the producers and the
point for which one wishes to find the field. Arrows should be drawn at that point, indicating the
direction of the field caused by each producer. The arrows are vectors and of course must be
added vectorially.
Example 11.1 p.548
Practice Problems p.548
Example 11.2 p.549
Practice Problems p.549
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Example 11.3 p.550
Practice Problems p.550
Example 11.4 p.551
Practice Problems p.551
Check and Reflect p.553 #1-9
MC #46-53
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Conductors and Electric Fields
For static equilibrium to exist, all charges must be at rest and therefore experience no net force.
In a conductor, electrons move freely about until they reach a state of static equilibrium. This
does not occur in non conductors.
Solid Conducting Sphere
All excess electrons on a solid conducting sphere repel each other and therefore distribute evenly
on the surface of the conducting sphere.
The electric field lines will always be perpendicular to the surface.
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Solid, Flat, Conducting Plate
The repulsive forces between the electrons are parallel to the surface. These forces depend on
how far apart each of the electrons are. For the forces to balance and the electrons to reach a
state of static equilibrium, the charges must be evenly distributed along the outer surface of the
plate.
The electric field lines will always be perpendicular to the plate.
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Irregularly Shaped Solid Conducting Object
On a flatter section of an irregularly shaped conducting object, the repulsive forces are parallel
to the surface such that the electrons spread out (similar to above). On a pointed part, the forces
act at an angle, so a smaller component is parallel. There is less repulsion, and therefore the
electrons collect closer together.
Hollow Conducting Object
The distribution of electrons is similar to that of the solid conducting sphere. Electrons are
repelled outward and distribute evenly on the outer surface. There is no excess charge on the
inner surface. There is also no electric field inside the conductor.
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Parallel Plate Capacitors
Charges spread out in a similar way to those on individual solid, flat, conducting plates. Because
of attractive forces between oppositely charged plates, the charges will accumulate on the inner
surfaces. The electric field lines are uniform, except near the edges.
Then, Now, and Future: Defibrillators Save Lives p.559 #1-3
Minds On: Faraday’s Ice Pail p.559
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Electric Potential Energy, Work, Electric Potential, and Electric
Potential Difference
In gravity, work done against gravity is equal to the change in gravitational potential energy.
E p  W
Similarly, work done in an electric field is the change in electric potential energy.
E p  W
Electric potential energy can also be defined as the energy stored in the system of two charges a
certain distance apart.
Example 11.5 p.561
Practice Problems p.561
Example 11.6 p.563
Practice Problems p.563
Suppose a test charge is placed between two parallel plates. It will naturally move toward one
plate and away from the other. Suppose, however, that it moves in the opposite direction. This
requires an external force, and work is therefore done:

E p  W  F d
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Electric potential is defined as the amount of electric potential energy per unit charge.
V 
Ep
q
When a charge moves in an electric field, it experiences a change in electric potential. This
change is called electric potential difference.
V  V final  Vinitial
V 
E p
q
electron volt (eV)- the energy gained or lost by an electron when passing through a potential
difference of 1.0 V.
1 eV=1.60 x 10-19 J
Example 11.7 p.565
Practice Problems p.565
Example 11.8 p.566
Practice Problems p.566
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Electric Field Between Parallel Plates
A third electric field equation is used to find the electric field between two charged parallel
plates. It is as follows:
 V
E 
d

where E is the electric field strength between two charged, parallel plates (N/C or V/m), ΔV is
the potential difference between the plates (ΔV), and Δd is the distance between the plates (m).
Note that the electric field between two charged, parallel plates is a uniform electric field, unlike
the electric field surrounding a charged sphere. The second and third electric field equations
may therefore be used when dealing with two charged parallel plates, however the first one may
not.
A comparison of the electric field equations:
EQUATION
Valid for Uniform
/ Non-Uniform

q
E  k 12
r

 Fe
E
q
 V
E 
d
non-uniform
uniform or nonuniform
Used
around a producer
for an electric
experienced by a
charge
yes
no
no
no
yes
yes (when there is a
charge between the
plates)
no
yes (when the charge
is between two
plates)
yes
uniform
field
between two
charged, parallel
plates
Example 11.9 p.568
Practice Problems p.568
Check and Reflect p.569 #2-13
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Charged Particles in Electric Fields
A charged particle in an electric field will experience a force that can be found using the

 Fe
equation given in the previous section ( E 
). It is also sometimes useful to describe the
q
energies of the charged particle, particularly if the charged particle is being accelerated across
the electric field.
If a charged particle is accelerating across an electric field (or potential difference- a problem
will usually mention only one or the other, but both really exist whenever one does), the law of
conservation of energy applies. Before a particle has been accelerated across the electric field
(or potential difference), it will possess electric potential energy ( E  qV …given to you on your
E
equation sheet as V 
). It may also possess kinetic energy if it is already moving before it
q
is accelerated. After it has been accelerated, it will possess kinetic energy because it will be
moving. It may also possess electrical potential energy if it has been accelerated across only part
of the electric field (or potential difference).
In any case, as a charged particle speeds up through an electric field its potential energy is being
converted to kinetic energy (as in the case of a ball speeding up in a gravitational field).
Conversely, if the charged particle is slowing down, kinetic energy is being converted to
potential energy (as in the case of a ball slowing down in a gravitational field).
Example 11.10 p.571
Practice Problems p.571
Example 11.11 p.572
Practice Problems p.572
Example 11.12 p.574
Practice Problems p.574
Check and Reflect p.575 #1-13
Unit Review MC #54-65
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Diploma Exam Review Questions
Basic Electricity
Use the following information to answer the next question.
32. After contact, the charge on sphere A is
A. 1.05 μC
B. 2.10 μC
C. 3.45 μC
D. 6.90 μC
33. The number of excess electrons on a ball that has a charge of –3.60 10–17 C, expressed in scientific notation, is a.bc 10d . The values of a,
b, c, and d are _____ , _____ , _____ , and _____ .
(Record all four digits of your answer in the numerical-response section on the answer sheet.)
Charging Objects
Use the following information to answer the next two questions.
34. The droplets leave the nozzle with a
A. negative charge caused by the movement of protons onto the needle
B. positive charge caused by the movement of electrons onto the needle
C. positive charge caused by the movement of protons onto the droplets
D. negative charge caused by the movement of electrons onto the droplets
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35. The charged droplets are kept from being blown off of the leaves by the wind because the charged droplets
A. gain electrons from the air and transfer them to the leaves
B. fall faster through the air because they have similar charges
C. induce an opposite charge on the leaves so they are attracted to them
D. repel each other and spread out, thus the effect of the wind is minimized
Use the following information to answer the next question.
36. The distribution of charge on the rod is
A. positive at end X and electrons move off the rod into the ground
B. negative at end X and electrons move off the rod into the ground
C. positive at end X and electrons move onto the rod from the ground
D. negative at end X and electrons move onto the rod from the ground
Use the following information to answer the next two questions.
37. The bottom of a thundercloud usually becomes negatively charged. Before lightning strikes, the charge of the ground directly beneath the
thundercloud will become
A. positive by induction
B. negative by induction
C. positive by conduction
D. negative by conduction
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38. During the downward lightning strike, the charge on the top of the tree becomes
A. negative by induction
B. negative by conduction
C. neutral by induction
D. neutral by conduction
Use the following information to answer the next question.
39. When particles are between grids X and Y, they are repelled by
A. grid X and each other, but are attracted to grid Y
B. grid Y and each other, but are attracted to grid X
C. grid X but are attracted to each other and grid Y
D. grid Y but are attracted to each other and grid X
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Coulomb’s Torsion Balance Experiment
Use the following information to answer the next question.
40. In order to determine the relationship between force and distance, Coulomb needed to
A. keep the magnitude of one charge constant
B. keep the magnitude of both charges constant
C. keep the distance between the charges constant
D. vary the magnitude of one charge while varying distance between the charges
41. Newton’s Law of Universal Gravitation has a mathematical relationship similar to the one developed by
A. Coulomb
B. Einstein
C. Lenz
D. Ohm
Coulomb’s Law (2 Charges)
Use the following information to answer the next question.
42. The magnitude of the electric force exerted by sphere A on sphere B after contact and separation is __________ N.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
43. A small object carrying a charge of 3.47 μC experiences an electric force of 7.22 10–2 N when placed at a distance, d, from a second,
identically charged object. The value of d is __________ m.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
Coulomb’s Law (3 Charges)
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Electric Fields (Qualitative)
Use the following information to answer the next question.
44. The types of charge present on X and Y are, respectively,
A. negative and negative
B. negative and positive
C. positive and negative
D. positive and positive
45. A scalar field differs from a vector field in that
A. a scalar field acts in only one direction
B. a vector field acts in only one direction
C. direction is irrelevant for a scalar field
D. direction is irrelevant for a vector field
Electric Field Problems
Use the following information to answer the next question.
46. The electric field at a point halfway between the particles is
A. zero
B. toward the left of the page
C. toward the top of the page
D. toward the right of the page
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Use the following information to answer the next two questions.
47. The strength of the electrical field induced in the gap of the spark plug is
A. 6.0 N/C
B. 6.0 x 103 N/C
C. 1.0 x 104 N/C
D. 1.0 x 107 N/C
48. The acceleration of the electrons across the gap of the spark plug, expressed in scientific notation, is a.b x 10cd m/s2 . The values of a, b, c, and
d are ____, ____, ____, and ____ .
(Record your four-digit answer in the numerical response section on the answer sheet.)
49. A point charge of magnitude 6.9 10–5 C produces an electric field of 1.0 103 N/C at point P. The distance from P to the charge is
A. 4.3 10–2 m
B. 2.1 10–1 m
C. 2.5 101 m
D. 6.2 102 m
50. The intensity and direction of the electric field produced by an alpha particle at a distance of 5.0 10–11 m from the particle is
A. 5.8 1011 N/C, toward the alpha particle
B. 5.8 1011 N/C, away from the alpha particle
C. 1.2 1012 N/C, toward the alpha particle
D. 1.2 1012 N/C, away from the alpha particle
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51. The magnitude of an electric field at a distance x from a point charge Q is 8.3 10–4 N/C. If the distance is increased to 3x and the charge is
reduced to Q/4 , then the magnitude of the electric field will be
A. 1.9 10–3 N/C
B. 3.7 10–4 N/C
C. 6.9 10–5 N/C
D. 2.3 10–5 N/C
Use the following information to answer the next question.
52. The magnitude of the net electric field at point P due to these four point charges is
A. 5.4 104 N/C
B. 4.5 104 N/C
C. 2.7 104 N/C
D. 0.0 N/C
Use the following information to answer the next question.
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53. An electric field of magnitude 7.17 104 N/C is maintained between the grids of the electrostatic precipitator. The distance between grids X
and Y is 5.60 cm. The potential difference across grids X and Y is
A. 1.28 106 V
B. 4.02 105 V
C. 1.28 104 V
D. 4.02 103 V
Charged Particles in External Magnetic Fields
Unit 2 Review Questions
54. The electric field between a positive point charge and a negative point charge represented by




55. Two charged objects experience a force of 18.0 N when they are placed 5.00 × 10 –2 m apart. If the charge on one object is 1.30 × 10–5 C, then
the charge on the other object is a.bc × 10–d C. The values of a, b, c, and d are _____ , _____ , _____ , and _____.
(Record all four digits of your answer in the numerical-response section on the answer sheet.)
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Use the following information to answer the next four questions.
56. The minimum electron speed necessary to ionize xenon atoms is
A. 2.66 × 10 31 m/s
B. 5.15 × 10 15 m/s
C. 4.25 × 10 12 m/s
D. 2.06 × 10 6 m/s
57. The electric potential difference across the electrified grids that is required to accelerate a xenon ion from rest to its exit speed is
A. 2.93 × 10–5 V
B. 1.26 × 10–3 V
C. 1.26 × 103 V
D. 4.71 × 1029 V
58. If all of the xenon propellant could be expelled in a single short burst, the change in the speed of the DS1 capsule after all the fuel has been
exhausted would be
A. 6.14 m/s
B. 7.16 m/s
C. 6.14 × 103 m/s
D. 7.16 × 103 m/s
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59. The physics principle that best describes the propulsion of the DS1 capsule is the Law of Conservation of
A. Charge
B. Energy
C. Momentum
D. Nucleon Number
60. The energy gained by a proton that moves through a potential difference of 1.0 V is
A. 1.0 J
B. 1.0 eV
C. 6.3 × 1018 J
D. 1.6 × 10–19 eV
Use the following information to answer the next three questions.
61. When the neutral pith ball is placed near the charged Van de Graaff generator, the pithball is attracted to the generator as a result of
A. induction
B. grounding
C. conduction
D. induction and grounding
62. The direction of the electrical force on the pith ball is
A. 
B. 
C. 
D. 
63. The magnitude of the electrical force exerted on the pith ball by the charged Van de Graaff generator is
A. 2.5 × 10–3 N
B. 2.3 × 10–3 N
C. 8.9 × 10–4 N
D. 8.4 × 10–4 N
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64. The magnitude of the force between two charged particles that are a fixed distance apart is 3.80 × 10 –4 N. If the distance between their centres
is exactly doubled, then the magnitude of the force between the particles, expressed in scientific notation, is a.bc × 10–d N. The values of a, b, c,
and d are _____ , _____ , _____ , and _____.
(Record all four digits of your answer in the numerical-response section on the answer sheet.)
65. The electric field strength 2.0 × 10–10 m from an alpha particle is
A. 7.2 N/C
B. 14 N/C
C. 3.4 × 1010 N/C
D. 7.2 × 1010 N/C
Electricity MC Answers
32
33
34
35
36
37
38
39
40
A
2252
B
C
D
A
B
A
B
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42
43
44
45
A
15.9
1.22
A
C
46
47
48
49
50
B
D
1818
C
D
51
52
53
54
55
D
D
D
B
3857
56
57
58
59
60
D
C
D
C
B
61
62
63
64
65
A
B
C
9505
D
47
Domain Theory of Magnetism
A ferromagnetic material is a material that can be magnetized. Iron, nickel, and cobalt are
examples of ferromagnetic materials. Every material, including ferromagnetic materials, has
electrons. In ferromagnetic materials, the spin of the electrons line up in small regions. We call
one of these small regions where electron spins line up a “domain”. Various domains within the
material can align in a given direction when exposed to an external magnetic field (if we
represent a domain with an arrow, the domains are aligned when the arrows are all pointing in
the same direction). When the domains become aligned, the material has become magnetized.
Ferromagnetic Material With Unaligned Domains
Ferromagnetic Material With Aligned Domains
Aligned domains in a magnet point from the South pole of the magnet to the North Pole. When a
magnet is cut in half, two magnets will be formed, each with a North pole and a South pole.
Some materials can retain their magnetism permanently, while others retain their magnetism for
only a short period of time. A permanent magnet is sometimes called a “hard magnet” while a
temporary magnet is sometimes called a “soft magnet”.
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Laws of Magnetism
Magnets have the following properties:
1. Like poles of magnets repel.
2. Unlike poles of magnets attract.
3. Both north and south poles of a magnet attract non-magnetized ferromagnetic materials. This
happens because the magnet temporarily aligns domains in the non-magnetized
ferromagnetic material.
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Magnetic Fields
A magnetic field is a line of magnetic influence around a permanent or temporary magnet. There
is a direction associated with the influence of a magnetic field, so a magnetic field is a vector
field. The direction of the magnetic field is defined in two ways:
(1) the direction in which a compass needle will point when placed in the magnetic field
(2) from north to south
Let’s compare magnetic fields to the other types of fields we’ve done so far this year:
SOURCE
DIRECTION
gravitational field
object with mass
attractive
(toward producer)
electric field
charged object
attractive / repulsive
(direction a positive
charge moves when
placed in the field, away
from positive and toward
negative)
magnetic field
ferromagnetic material with
aligned domains
attractive / repulsive
(direction a compass
points, from North to
South)
Now let’s look at some diagrams that show the interaction of magnetic fields. As with electric
fields, parallel lines indicate a uniform field while non-parallel lines indicate a non-uniform field.
a.
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b.
c.
(Pearson Physics, Ackroyd et al, 2009)
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The Earth’s Magnetic Field
If magnetic field lines point from north to south and in the direction a compass points when
placed in the field, then a compass points toward the south. Does this mean that the South Pole
is somewhere above Canada? Absolutely. An implication of the definitions of the direction of
magnetic fields is that the “North Pole” of Earth is not truly a magnetic “North Pole”, rather it is
actually a magnetic “South Pole”. Examine the diagram below, keeping in mind both definitions
of magnetic field direction.
Then, Now, and Future: Earth’s Magnetic Field p.591 #1-3
MC #66-68
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Magnetism From Electricity
In 1819, Hans Christian Oersted discovered a relationship between electricity and magnetism.
Oersted found that an electric current in a wire could be deflected a nearby compass needle. He
thus concluded that electric currents (or simply moving charges) produce magnetic fields.
Oersted’s discovery tells us that any moving electric charge produces a magnetic field. It tells us
nothing of the magnitude or direction of that magnetic field. The direction of the magnetic field
produced by a moving negative charge (or electron current) can be found by the Wire Grasp
Rule. To use this rule for negative charges, you stick your left thumb in the direction that the
negatively charged particle is moving, then clench your fist. Your fingers point in the direction
of the magnetic field produced by the moving negative particle. Notice that as you turn your
hand, the direction in which your fingers are pointing changes. This results in a magnetic field
that actually forms a circle around the moving negative particle. If the field around a moving
positively charged particle is to be found, also use the Wire Grasp Rule but use your right hand
instead of your left.
(Pearson Physics, Ackroyd et al, 2009)
Some Examples:
magnetic
field
magnetic
field
current flow
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current flow
magnetic field
(into page)

current flow

magnetic field
(out of page)
53


current into page
magnetic field CCW
current out of page
magnetic field CW
You can, of course, find the direction of the current given the direction of the magnetic field that
caused the current by performing the rule in reverse: clench your fist with your fingers pointing
in the direction of the magnetic field then stick out your thumb. Your thumb will be pointing in
the direction of the current.
If the current is moving through a solenoid (a coil of wire), you can use a slightly different
version of the wire grasp rule (we will call it the Newfangled Coil Rule) to find the direction of
the magnetic field. In this case, simply clench the solenoid with your hand, with your clenched
fingers pointing in the direction of the current. Your thumb will automatically point in the
direction of the induced magnetic field.
(Pearson Physics, Ackroyd et al, 2009)
An Example Not in Your Book
Draw the magnetic field surrounding the electron currents below:
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Another Example Not in Your Book
Draw the magnetic field surrounding the electron currents below:
Yet Another Example Not in Your Book
Draw the magnetic field surrounding the electron currents below:
Check and Reflect p.592 #2,4-8,10-16
Skills Practice p.588 #1
MC #69-71
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Charged Particles in External Magnetic Fields (The Motor Effect)
When a positively charged particle (whether moving or at rest) is exposed to an external electric
field, a force is exerted on the particle in the direction of the external field. Thus, the positively
charged particle will deflect in the direction of the external electric field. When a negatively
charged particle (whether moving or at rest) is exposed to an external electric field, a force is
exerted on the particle in the direction opposite to that of the external field. Thus, the negatively
charged particle will deflect in the opposite direction of the electric field.
A negatively or positively charged particle at rest will not be deflected by an external magnetic
field, however one in motion will. The direction of the deflection can be found by using the
Hand Rule For Deflection. For negatively charged particles, use the Left Hand Rule For
Deflection. The first step is to extend your fingers on your left hand and point them in the
direction of the external magnetic field. Your left thumb should be pointing at a right angle to
your extended fingers, and it should be pointed in the direction that the negatively charged
particle is traveling. The palm of your left hand will be pointing in the direction of the magnetic
force. The deflection of the particle will be in the direction of the magnetic force. For a moving
positively charged particle, of course, the Right Hand Rule For Deflection applies. Again, it is
identical to the left hand rule, except that the right hand is used in place of the left. Note that this
rule assumes that the particle is moving perpendicular to the magnetic field.
(Pearson Physics, Ackroyd et al, 2009)
The magnitude of the magnetic force acting on a moving charged particle is given by the
following equation:


F  qv B
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where F is the magnitude of the magnetic force (N), q is the magnitude of the charge (C), v is the
speed with which the particle is traveling (m/s), and B is the strength of the external magnetic
field (T).
Note that the above equation assumes that the particle is moving perpendicular to the field. This
results in the maximum possible magnetic force. If the particle is moving parallel to the field,
the magnetic forces is zero. If the angle is something other than 00, 900, or 1800, the magnetic
force will have a value somewhere between 0 N and the maximum force.
You can, of course, find the direction of the particle given the direction of the magnetic field that
and force by performing the rule in reverse: Outstretch your fingers. Point them in the direction
of the external magnetic field. Point your palm in the direction of the magnetic force. Your
fingers should point in the direction that the particle is moving.
Example 12.1 p.599
Practice Problems p.599 (#1 and #3)
Example 12.2 p.600
Practice Problems p.600
MC #72-74
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More on Charged Particles in External Magnetic Fields
A charged particle moving through a uniform external magnetic field will travel in a circular
path until it exits the uniform magnetic field. Because it travels in a circle, there is a centripetal
force, which is caused by the magnetic force. The centripetal force and magnetic force are
therefore equal:
FC  FM
m
v2
 qv B
r
where m is the mass of the charged particle (kg), v is the speed of the charged particle (m/s), r is
the radius of curvature of the circular path of the particle (m), q is the charge of the particle (C),
and B is the strength of the uniform external magnetic field (T).
Whether a particle curves up or down depends on the sign of the particle and the direction of the
external magnetic field. Its direction of curvature can be found using the Hand Rule For
Deflection.
An Example Not in Your Book
An electron is moving with a speed of 4.0 x 105 m/s to the right through a magnetic field of
magnitude 2.5 x 10-2 T directed into the page. What is the radius of the path of the electron and
in which direction will it bend (up/down)?
Check and Reflect p.601 #3-10
MC #75-84
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Electric Current
Electric current is rate of flow of electric charge.
I
q
t
The unit for electric charge is the Ampere (A).
Example 12.3 p.603
Practice Problems p.603
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Current Carrying Conductors in Magnetic Fields
A conductor is simply a wire that allows electrons to flow through it. Because the flowing
electrons are moving negatively charged particles, there is a magnetic force on the currentcarrying conductor in the same direction as it would be on moving electrons.
The direction of the magnetic force on a current-carrying conductor can be found by using the
Hand Rule For Magnetic Force. The Hand Rule For Magnetic Force is virtually identical to the
Hand Rule For Deflection, except that the thumb is pointed in the direction of the current instead
of the single moving charge. Essentially, you can ignore the fact that a wire even exists and treat
the problem as simply a moving charged particle in an external magnetic field. Again, this rule
assumes that the direction of the current is perpendicular to the magnetic field.
(Pearson Physics, Ackroyd et al, 2009)
The magnitude of the magnetic force acting on a current-carrying conductor is given by the
following equation:
F  Il B
where F is the magnitude of the force (N), I is the strength of the current in the conductor (A), l
is the length of the current-carrying conductor (m), and B is the strength of the external magnetic
field (T).
Remember that the conductor must have a current flowing through it, and that the above equation
does not require the conductor itself to be moving.
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If a conductor is suspended in a magnetic and gravitational field, the forces caused by these
fields must balance. The following equation can therefore be used:
Fg  FM
mg  Il B
Example 12.2 p.605
An Example Not in Your Book
A 0.100 m long current-carrying wire with a mass of 2.00 x 10-1 kg is exposed to a magnetic
field. What is the magnetic field strength required to suspend the wire if the current through the
wire is 2.50 x 10-1 A?
Practice Problems p.605
MC #85-86
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Summary of Left (and Right) Hand Rules
Wire Grasp Rule:
direction of current
THUMB
direction of magnetic field caused by current
CLENCHED FINGERS
Hand Rule For Deflection:
direction of current
THUMB
direction of external magnetic field
OUTSTRETCHED FINGERS
direction of magnetic force
PALM
Hand Rule For Magnetic Force:
direction of current
THUMB
direction of external magnetic field
OUTSTRETCHED FINGERS
direction of magnetic force
PALM
An Example Kind of In Your Book:
Using the diagram below, describe the operation of the electric motor.
(Pearson Physics, Ackroyd et al, 2009)
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Moving a Conductor Through A Magnetic Field (The Generator
Effect)
In 1831, Michael Faraday discovered what amounts to the converse of Oersted’s discovery.
Faraday discovered that when a wire is moved relative to a magnet (or a magnet is moved
relative to a wire), or a wire is exposed to a magnetic field that is changing in magnitude, then a
potential difference and therefore an electric current is induced in the wire. This is known as the
generator effect.
Because a conductor is made up of electrons (and other particles as well), it can be treated the
same as moving an electron in an external magnetic field. An electron moving in an external
magnetic field experiences a magnetic force. The electrons in a conductor that is moving in an
external magnetic field experience a magnetic force as well. These electrons experiencing the
force start to move, resulting in an electric current in the conductor.
(Pearson Physics, Ackroyd et al, 2009)
The direction of the induced current can be found by the Hand Rule For Deflection. If, as in the
example above, the wire is moving upwards, then electrons within the wire are also moving to
the upwards. You would therefore stick your thumb upwards. Because the external magnetic
field is to the right (from North to South), you would stick your outstretched fingers to the right.
Your palm will point in the direction of the magnetic force (on the electron) and therefore the
direction of the current in the wire.
Check and Reflect p.613 #1-12
Unit Review MC #88-96
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Diploma Review Questions
Use the following information to answer the next question.
66. Given the magnetic fields illustrated above, the magnets will repel in diagrams
A. I and II only
B. II and III only
C. I and IV only
D. II and IV only
-----------------------------------------------------------------Use the following information to answer the next question.
67. The two bar magnets cause the net magnetic field at P to be in the direction
A. east
B. west
C. north
D. south
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Use the following information to answer the next question.
68. The direction of the magnetic field at P due to the two bar magnets is
Magnetic Fields Produced by Electric Current
Use the following information to answer the next question.
69. The compass that correctly indicates the direction of the magnetic field produced by a wire conducting electrons is numbered
A. 1
B. 2
C. 3
D. 4
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70. An automobile’s battery delivers a steady DC current to a headlight. The electric current in the wire produces a circular
A. electric field around the wire
B. magnetic field around the wire
C. gravitational field around the wire
D. electromagnetic field around the wire
Use the following information to answer the next question.
71. What is the direction of the magnetic field at point P?
A. Into the page
B. Out of the page
C. Toward the top of the page
D. Toward the bottom of the page
Moving Charges in External Magnetic Fields
Use the following information to answer the next question.
72. If the electron is deflected downward in each field, then field 1, field 2, and field 3 are, respectively,
A. electric, magnetic, and gravitational
B. gravitational, magnetic, and electric
C. magnetic, gravitational, and electric
D. magnetic, electric, and gravitational
------------------------------------------------------------------
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73. The path followed by a moving proton in an external magnetic field is shown in
74. The magnitude of the magnetic force exerted on a charged particle in a magnetic field will be doubled by doubling any one of
A. the charge of the particle, or the speed of the particle, or the mass of the particle
B. the magnitude of the field or the angle of entry of the particle
C. the speed of the particle, or the mass of the particle, or the magnitude of the field
D. the charge of the particle, or the speed of the particle, or the magnitude of the field
Use the following information to answer the next question.
75. The protons in the solar wind experience a magnetic force
A. into the plane of the page
B. out of the plane of the page
C. in the direction the solar wind is traveling
D. opposite to the direction the solar wind is traveling
76. Assume that the velocity of the solar wind particles is perpendicular to the magnetic field. The radius of the circular path that protons in a
solar wind follow, expressed in scientific notation, is a.bc x 10d m. The values of a, b, c, and d are ____, ____, ____, and ____.
(Record your four-digit answer in the numerical response section on the answer sheet.)
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Use the following information to answer the next question.
77. The curved paths of the particles in the pair production diagram result from the electron and positron moving through an external magnetic
field. In this diagram, the direction of the magnetic field causing these paths to curve is
A. into the page
B. out of the page
C. to the left
D. to the right
78. During pair production, the speed of the electron or of the positron can be calculated by measuring the radius of the circular path it travels
within the magnetic field. The speed of a charged particle moving in a circular path in a
uniform magnetic field is given by
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Use the following information to answer the next question.
79. The coils that can produce a deflection toward the top of the screen are numbered
A. 1 and 3
B. 2 and 4
C. 1 and 2
D. 3 and 4
Use the following information to answer the next two questions.
80. The speed of the proton is
A. 4.14 105 m/s
B. 1.77 107 m/s
C. 1.71 1011 m/s
D. 3.14 1014 m/s
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81. The strength of the magnetic field, expressed in scientific notation, is a.bc 10d T. The values of a, b, c, and d are _____ , _____ , _____ ,
and _____ .
(Record your four-digit answer in the numerical-response section on the answer sheet.)
82. A proton and an alpha particle have identical circular orbits in a magnetic field. The proton has a speed of 4.4 105 m/s. The speed of the
alpha particle is
A. 1.1 105 m/s
B. 2.2 105 m/s
C. 4.4 105 m/s
D. 8.8 105 m/s
Use the following information to answer the next two questions.
83. The maximum speed of the protons in Lawrence’s cyclotron was
A. 1.5 1013 m/s
B. 1.7 108 m/s
C. 3.9 106 m/s
D. 9.8 1015 m/s
84. The magnitude of the magnetic field used by Lawrence was
A. 6.3 10–1 T
B. 2.7 101 T
C. 2.4 106 T
D. 1.6 109 T
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Electric Current in an External Magnetic Field
85. A 50.0 cm length of wire has a weight of 0.389 N and a current of 0.250 A. The wire remains suspended when placed perpendicularly across
a magnetic field. The strength of the magnetic field is __________ T.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
86. A wire that is 75.0 cm long carries a current of 6.00 A. The wire is at right angles to a uniform magnetic field and experiences a magnetic
force of 0.350 N. The magnitude of the magnetic field, expressed in scientific notation, is b 10–w T. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
87. XXXX
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Magnetism Review Questions
Use the following information to answer the next three questions.
88. The statements that describe the motion of the charged particle in diagram 1 are
A. I and III
B. I and IV
C. II and III
D. II and IV
89. The statements that describe the motion of the charged particle in diagram 2 are
A. I and III
B. I and IV
C. II and III
D. II and IV
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90. The direction of the uniform magnetic field in diagram 2 is
A. toward the top of the page
B. toward the bottom of the page
C. to the left of the page
D. to the right of the page
Use the following information to answer the next question.
91. The magnetic field induced around the rubber rod as it moves is represented by
92. An alpha particle travels at 1.08 × 105 m/s perpendicularly through a magnetic field of strength 1.12 × 10 –3 T. The magnitude of the magnetic
force on the alpha particle is b × 10–w N. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
93. A copper wire is connected to a battery so that it has a current in it. A segment of the wire is perpendicular to a horizontal 1.5 T magnetic
field. The length of the wire in the magnetic field is 3.0 cm, and the mass of the wire affected by the magnetic field is 20 g. In order to suspend
the segment of wire, the minimum current in the wire must be
A. 0.044 A
B. 0.23 A
C. 4.4 A
D. 44 A
Use the following information to answer the next question.
94. A possible explanation for the deflection of the compass needle is that the
A. bottom of the filing cabinet is positively charged
B. bottom of the filing cabinet is negatively charged
C. induced magnetic polarity of the bottom of the filing cabinet is opposite to that at the top of the filing cabinet
D. bottom of the filing cabinet is closer to Earth so it is more strongly magnetized than the top of the filing cabinet
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95. If the source of Earth’s magnetic field were a bar magnet, then the best diagram to show this field would be
Use the following information to answer the next question.
96. The ratio of radius of the alpha particle’s path to the radius of the proton’s path is
A. 1 : 1
B. 2 : 1
C. 4 : 1
D. 8 : 1
Magnetism MC Answers
66
67
68
69
70
D
B
D
C
B
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72
73
74
75
B
A
A
D
B
76
77
78
79
80
7594
A
A
B
A
81
82
83
84
85
1201
B
C
A
3.11
86
87
88
89
90
7.78
A
C
B
B
91
92
93
94
95
96
C
3.87
C
C
D
C
74
Electromagnetic Radiation
Maxwell knew that a changing magnetic field produced a changing electric field. He
hypothesized that, by symmetry, a changing electric field would produce a changing magnetic
field, and so on. Electromagnetic radiation is caused by accelerating electric charges. Note that
this can include a charged particle speeding up or slowing down. Maxwell calculated the speed
of these electromagnetic waves to be 3.00 x 108 m/s.
Electromagnetic radiation is a transverse wave consisting of a sinusoidal electric field that is
perpendicular to, and in phase with, a sinusoidal magnetic field. The direction of propagation of
the wave is perpendicular to both fields.
These electromagnetic waves propagate (travel) at the speed of light (3.00 x 108 m/s in a
vacuum).
Electromagnetic radiation exhibits the following properties of waves: polarization, diffraction,
interference, and refraction.
Heinrich Hertz was the first to experimentally detect electromagnetic radiation. He detected low
frequency electromagnetic radiation using a spark-gap apparatus. He named the electromagnetic
radiation that he detected radio waves.
With the spark-gap apparatus, Hertz used a step-up transformer to accelerate electrons across a
gap in the secondary coil. This acceleration of electrons produced electromagnetic radiation that
was detected some distance away by using a type of antenna.
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(Pearson Physics, Ackroyd et al, 2009)
Check and Reflect p.647 #1-2, 5,7-9, 11-14
MC #97-105
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Determining the Speed of EMR
The currently accepted value for the speed of EMR in air and in a vacuum is approximately 3.00
x 108 m/s (Maxwell had predicted 3.10 x 108 m/s). The following describes many early attempts
to experimentally arrive at the speed of EMR.
Galileo
Galileo and an assistant stood on hills 1 km apart. Galileo uncovered a lantern. When his
assistant saw the light, he uncovered a lantern. Galileo measured the time between when he
uncovered his lantern and when he saw the light from the second lantern return, and from this,
calculated the speed of light. Galileo was not at all close in his determination of the speed of
EMR because the high speed of the EMR made the reaction time involved in his experiment
much too significant.
Romer and Hugens
The period of revolution of Io (Jupiter's brightest moon) was known. Roemer knew that Io
should be eclipsed (disappear behind Jupiter) every 42.5 hours.
Roemer discovered that Io took longer to be eclipsed when Earth was farther from Jupiter and
less time when Earth was closer to Jupiter. The time between the extremes was 22 minutes.
(Pearson Physics, Ackroyd et al, 2009)
The 22 minutes ends up being the time that it takes light to travel Earth's orbital diameter. From
this, Roemer could calculate the speed of light.
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Fizeau
Light was directed from a source to a mirror 8.63 km away. It needed to pass through an
opening in a toothed wheel to get there. The light reflected off of the mirror and back to an
observer. It again needed to pass through the toothed wheel.
Knowing the frequency of rotation of the toothed wheel, Fizeau could determine the time that it
took the light to travel the 17.26 km round trip. From this information, he could calculate the
speed of light (3.15 x 108 m/s).
(Pearson Physics, Ackroyd et al, 2009)
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Michelson
Michaelson directed light off of a rotating 8-sided mirror on one hill to another hill 35 km away.
The light reflected off of a mirror on the second hill, and back to the octagonal rotating mirror on
the first hill.
(Pearson Physics, Ackroyd et al, 2009)
If the rotating mirror made 1/8 (or a multiple or 1/8) of a revolution during the time of the light's
travel, the light would be reflected to an observer. Knowing the frequency of rotation of the
octagonal mirror, Michelson could determine the time that it took the light to travel the 70 km
round trip. From this, Michelson could determine the speed of light.
Example 13.1 p.650
Another Example Not in Your Book
Students performed an experiment similar to Michelson’s with a 10 sided rotating mirror located
30.0 km from a mirror on a second hill. The frequency of the 10 sided mirror is adjusted until
light is observed by the observer. At what frequency does the observer see the reflected light?
Practice Problems p.650
Check and Reflect p.652 #1-11
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The Law of Reflection
Recall from your waves unit that the law of reflection states that the angle of incidence of a wave
is equal to the angle of reflection of a wave. Since both waves and particles reflect and obey the
law of reflection when they encounter a new medium, light could be thought of as either a wave
or a particle in the context of reflection.
To make it easier to visualize the law of reflection, we often represent a light wave or particle’s
motion by a ray. A ray is a straight line with an arrow on the end of it that represents the
rectilinear propagation of the light.
An Example Not in Your Book:
Assume the solid lines below are mirrors. Draw the reflected ray(s) for each diagram and
calculate the angle(s) of reflection:
There are actually two types of reflection: specular and diffuse. Both obey the law of reflection.
(Pearson Physics, Ackroyd et al, 2009)
Specular reflection occurs when light is reflected from a smooth surface and all parallel incident
rays reflect parallel to each other. Diffuse reflection occurs off of rough surfaces and causes
parallel incident rays to reflect in seemingly random directions (note that the law of reflection is
still obeyed for each ray). Diffuse reflection allows us to see reflected light from all angles,
whereas specular reflection requires us to be in a specific spot to see the reflected light.
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Curved Mirrors (Ray Diagrams)
The two main types of curved mirrors are concave and convex mirrors. Light rays reflect and
converge when they strike a concave mirror, while they reflect and diverge when they strike a
convex mirror.
The following is a diagram of a concave mirror, with some of its characteristics labelled:
object
centre of curvature focal point
(C)
(f)
principal axis (PA)
Draw three rays on the above diagram:
1. From the object to the mirror, parallel to the principal axis. This ray reflects down through
the focal point.
2. From the object through the focal point. This ray reflects back parallel to the principal axis.
3. From the object through the centre of curvature. This ray reflects back along the same path.
Now draw an arrow from the principal axis to where the lines intersect. This arrow represents
the image.
Note that for any given ray diagram, you need to draw only two of these rays.
There are two types of images that can be formed: real and virtual.
A real image is an image that can be projected onto a screen. If an image formed is a real
image, you can tell from your ray diagram by either of two observations: if the image is
formed on the same side of the mirror as the object, the image is real, and if the image is
formed from two solid lines intersecting, the image is real.
A virtual image is an image that cannot be projected onto a screen. A person viewing a
virtual image must physically look at the mirror in order to see it. If an image formed is
virtual, you can tell from your ray diagram by either of two observations: if the image is
formed on the opposite side of the mirror as the object, the image is virtual, and if the
image is formed from one or more dotted lines (backwards extensions of the solid lines)
intersecting, the image is virtual.
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Each image formed has three characteristics:
1) Attitude: If the image is on the same side of the principal axis as the object, we say it is
erect but if it is on the other side, we say it is inverted.
2) Magnification: If the image is larger than the object, we say it is enlarged, but if it is
smaller than the object, we say it is diminished.
3) Type: If the image is drawn at the point where solid lines intersect, we say the image is
real. If it is drawn where extensions of solid lines (dotted lines) intersect, we say the
image is virtual.
The image formed in the ray diagram above is smaller, inverted, and real.
The following is a diagram of a convex mirror, with some of its characteristics labelled:
object
focal point centre of curvature
(f)
(C)
principal
axis
(PA)
Draw three rays on the above diagram:
1. From the object to the mirror, parallel to the principal axis. This ray reflects away from the
focal point. Draw a dotted extension of the reflected ray through the mirror all the way to the
focal point.
2. From the object toward the focal point. This ray reflects back parallel to the principal axis.
Draw a dotted extension of the reflected ray through the mirror parallel to the principal axis.
3. From the object toward the centre of curvature. This ray reflects back along the same path.
Draw a dotted extension of the reflected ray through the mirror all the way down to the
centre of curvature.
Now draw an arrow from the principal axis to where the lines intersect. This arrow represents
your image.
Note that for any given ray diagram, you need to draw only two of these rays.
The image formed in the ray diagram above is smaller, inverted, and real.
An Example Not in Your Book
For the diagrams on the following two pages, do the following:
1. Draw the ray diagram.
2. Identify the type of mirror and the characteristics of the image.
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Curved Mirrors (Mathematics)
The distance from the mirror to the focal point is called the focal length. Its symbol is f. Its
typical units are cm. The focal length of the mirror is one half of the radius of curvature (r).
The distance from the mirror to the object is called the object distance. Its symbol is do. Its
typical units are cm.
The distance from the mirror to the image is called the image distance. Its symbol is di. Its
typical units are cm.
The height of the object is called the object height. Its symbol is ho. Its typical units are cm.
The height of the image is called the image height. Its symbol is hi. Its typical units are cm.
The relative size of an image to an object is called magnification. Its symbol is M. It has no
units.
The following table shows the signs that are given to the above variables when using them to
solve problems:
f
dO
di
hO
hI
REAL
positive
positive
positive
VIRTUAL
negative
negative
negative
ERECT
INVERTED
positive
positive
negative
negative
“m” is positive when the image is on the same side as the object, and it is negative when it is
on the opposite side as the object.
The equation that relates together focal length, object distance, and image distance of a curved
mirror system is the following:
1
1
1


f do di
The equation that relates together magnification, image height, object height, image distance, and
object distance of a curved mirror system is the following:
m
hi
d
 i
ho
do
Example 13.2 p.664
Practice Problems p.664
Check and Reflect p.665 #1,3-11
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Refraction of Light
Think back to your physics 20 waves unit. When a wave passes from one medium to another, its
wavelength and speed both change. If the wave encounters the boundary between the two media
at an angle other than straight on, the change in speed causes the wave to change direction. This
change in direction of the wave is called refraction.
Refraction is a property of waves but not of particles. According to the law of refraction, light
must therefore be a wave.
The ratio of the speed of light in a vacuum (3.00 x 108 m/s) to the speed of light in a particular
medium is called the index of refraction, or refractive index, of the medium. The higher the
index of refraction of a medium, the more light will bend when it passes through the medium.
The symbol used to represent the index of refraction is n. It has no units. It can be found using
the following equation:
speed of light in vacuum
n=
speed of light in medium
Every material has a different index of refraction, and even within one particular material, there
is a different index of refraction for each different frequency (colour) of light. Since each
frequency results in a different index of refraction, different frequencies of light bend at different
angles in the same medium. The effect of this is that white light will split up into its component
colours (red, orange, yellow, green, blue, indigo, and violet) when it passes through the material.
This phenomenon is called dispersion.
The following diagram shows the phenomenon of dispersion, where the incident white light
splits up into its component colours. Red light has the lowest frequency and therefore refracts
the least, while violet light has the highest frequency and therefore refracts the most.
(Pearson Physics, Ackroyd et al, 2009)
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(Pearson Physics, Ackroyd et al, 2009)
When refraction occurs, the relationship between the angle of incidence and the angle of
refraction, the speed in the first material and the speed in the second material, the wavelength in
the first material and the speed in the second material, and the indices of refraction of the first
and second materials was discovered by Snell. The relationship is called Snell’s Law and can be
expressed mathematically by the following equation:
sin  1 v1 1 n2



sin  2 v2  2 n1
where 1 is the angle of incidence ( 0 ), 2 is the angle of refraction ( 0 ), v1 is the speed of light in
the first medium (m/s), v2 is the speed of light in the second medium (m/s), 1 is the wavelength
of light in the first medium (m), 2 is the wavelength of light in the second medium, n1 is the
index of refraction in the first medium, and n2 is the index of refraction in the second medium.
Note that in the above equation, only two of the parts ever need to be used.
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Total internal reflection occurs when light rays coming from the incident medium reflect at the
boundary, rather than refract. There is a minimum incident angle at which this will occur. This
minimum angle is called the critical angle. An incident ray that is at the critical angle is called
the critical ray. In order for total internal reflection to occur, light must be coming from a
medium with a high index of refraction to a medium with a low index of refraction, and the angle
of incidence must be greater than the critical angle. To find the value of the critical angle, you
should use Snell’s Law and set the angle of refraction equal to 900.
Example 13.3 p.668
Practice Problems p.668
Example 13.4 p.670
Practice Problems p.670
Example 13.5 p.673
Practice Problems p.673
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Curved Lenses (Ray Diagrams)
The two main types of curved lenses are concave and convex lenses. Light rays reflect and
converge when they strike a convex lens, while they reflect and diverge when they strike a
concave lens.
The following is a diagram of a convex lens, with some of its characteristics labelled:
Object
optical centre
focal point
(f)
principal axis (PA)
Draw two rays on the above diagram:
1. From the object to the mirror, parallel to the principal axis. This ray refracts down through
the focal point.
2. From the object through the optical centre of the lens. This ray keeps going in a straight line.
Now draw an arrow from the principal axis to where the lines intersect. This arrow represents
the image.
There are two types of images that can be formed: real and virtual.
A real image is an image that can be projected onto a screen. If an image formed is a real
image, you can tell from your ray diagram by either of two observations: if the image is
formed on the opposite side of the lens as the object, the image is real, and if the image is
formed from two solid lines intersecting, the image is real.
A virtual image is an image that cannot be projected onto a screen. A person viewing a
virtual image must physically look into the lens in order to see it. If an image formed is
virtual, you can tell from your ray diagram by either of two observations: if the image is
formed on the same side of the lens as the object, the image is virtual, and if the image is
formed from one or more dotted lines (backwards extensions of the solid lines)
intersecting, the image is virtual.
Each image formed has three characteristics:
1) Attitude: If the image is on the same side of the principal axis as the object, we say it is
erect but if it is on the other side, we say it is inverted.
2) Magnification: If the image is larger than the object, we say it is enlarged, but if it is
smaller than the object, we say it is diminished.
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3) Type: If the image is drawn at the point where solid lines intersect, we say the image is
real. If it is drawn where extensions of solid lines (dotted lines) intersect, we say the
image is virtual.
The image formed in the ray diagram above is smaller, inverted, and real.
The following is a diagram of a concave lens, with some of its characteristics labelled:
optical centre
object
focal point
(f)
principal
axis
(PA)
Draw two rays on the above diagram:
1. From the object to the lens, parallel to the principal axis. This ray refracts away from the
focal point. Draw a dotted extension of the reflected ray through the lens all the way to the
focal point.
2. From the object through the optical centre of the lens. This ray keeps going in a straight line.
Now draw an arrow from the principal axis to where the lines intersect. This arrow represents
your image.
The image formed in the ray diagram above is smaller, upright, and virtual.
An Example Not in Your Book:
For the diagrams on the following two pages, do the following:
1. Draw the ray diagram.
2. Identify the type of lens and the characteristics of the image.
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Type of lens: _______________
Type of lens: _______________
Type of lens: _______________
Type of lens: _______________
Type of lens: _______________
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Type of lens: _______________
Type of lens: _______________
Type of lens: _______________
Type of lens: _______________
Type of lens: _______________
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Curved Lenses (Mathematics)
The distance from the lens to the focal point is called the focal length. Its symbol is f. Its typical
units are cm.
The distance from the lens to the object is called the object distance. Its symbol is do. Its typical
units are cm.
The distance from the lens to the image is called the image distance. Its symbol is di. Its typical
units are cm.
The height of the object is called the object height. Its symbol is ho. Its typical units are cm.
The height of the image is called the image height. Its symbol is hi. Its typical units are cm.
The relative size of an image to an object is called magnification. Its symbol is M. It has no
units.
The following table shows the signs that are given to the above variables when using them to
solve problems:
f
dO
di
hO
hI
REAL
positive
positive
positive
VIRTUAL
negative
negative
negative
UPRIGHT
INVERTED
positive
positive
negative
negative
“m” is positive when the image is on the same side as the object, and it is negative when it is
on the opposite side as the object.
The equation that relates together focal length, object distance, and image distance of a curved
lens system is the following:
1
1
1


f do di
The equation that relates together magnification, image height, object height, image distance, and
object distance of a curved lens system is the following:
m
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d
 i
ho
do
93
Example 13.7 p.681
Practice Problems p.681
Check and Reflect p.683 #1-2, 5, 7-21
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Diffraction and Interference of Light
Recall from your waves unit that waves can diffract (spread out) when they go through an
opening. Also recall that waves can interfere either constructively or destructively when they
pass through each other.
In 1801, Thomas Young performed an experiment that demonstrated that light will diffract when
it goes through a narrow opening, and will interfere when it passes through other light. The
experiment, called Young’s Double Slit Experiment, thus demonstrated that light behaves like,
and therefore must be, a wave. Young used this experiment to measure the wavelengths of
visible light.
Young’s experiment was set up as follows:
Light waves pass through two narrow openings and diffract. Where waves intersect, there will
be areas of constructive or destructive interference. The screen on the bottom will detect these
areas of interference.
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A graph of the intensity of the interference pattern as a function of the position on the screen
follows:
Intensity
2nd order maximum
2nd order minimum
1st order maximum
1st order minimum
central maximum
1st order minimum
1st order maximum
2nd order minimum
2nd order maximum
x
Two equations can be used to mathematically relate the wavelength of the light to the
interference pattern observed. They are as follows:

d sin 
n
where λ is the wavelength of the light passing through the slits (m), d is the distance between the
slits (m), Ө is the angle between the centre of the central maximum and the centre of the nth
maximum ( 0 ), n is the number of maxima encompassed by the angle θ (no units- measured from
the central maximum),
and

xd
nl
where λ is the wavelength of the light passing through the slits (m), d is the distance between the
slits (m), x is the distance between maxima (m), l is the distance from the slits to the screen (m),
n is the number of maxima encompassed by the distance x (no units).
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Note that the first equation can only be used when measurements are taken from the central
maximum, while the second equation can be used in any situation.
Sometimes a diffraction grating (more than two slits) is used in this experiment. Typically, the
number of lines (or slits) per metre will be given to you. This information can be converted to d
by taking its inverse, then used in one of the above equations.
Example 13.8 p.689
Practice Problems p.689
Example 13.9 p.691
Practice Problems p.691
Example 13.10 p.693
Practice Problems p.693
Polarization
Polarization is a process that uses one or more filters to prevent one or more components of a
wave from passing through. Light can be thought of as an electromagnetic wave, with an electric
field wave and a magnetic field wave. A polarizing filter can be used to allow electric field
waves only on one plane (and therefore their corresponding magnetic field waves) through. If
two polarizing filters are used perpendicular to each other, the entire electromagnetic wave can
be blocked.
(Pearson Physics, Ackroyd et al, 2009)
Check and Reflect p.697 #6-11
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The Photoelectric Effect
In 1887, Heinrich Hertz was performing experiments testing the electromagnetic theory of waves
(recall that Hertz was the first to experimentally detect electromagnetic radiation). Hertz noticed
that certain high frequency (low wavelength) light striking certain metallic surfaces caused
electrons to be ejected. This effect became known as the photoelectric effect because the effect
required light to eject electrons. The ejected electrons came to be known as photoelectrons.
Hertz hypothesized that higher intensity light would cause more electrons to be emitted. This
hypothesis, however, is not correct. The correct explanation of the photoelectric effect came
from Albert Einstein, when he used his photon theory to explain it. Einstein defined a photon as
a quantum of EMR. It essentially means that light is a particle, with each particle of light having
c
a certain characteristic energy. Its energy is related to its frequency ( E  hf  h ).

Example 14.1 p.706
Practice Problems p.706
Example 14.2 p.707
Practice Problems p.707
Example 14.3 p.708
Practice Problems p.708
Check and Reflect p.710 #1-6
MC #106-113
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Einstein’s Explanation of the Photoelectric Effect
Intensity of EMR in photon theory is recognized as being the number of photons, not the energy
of the light. The higher the intensity of light, the more photons it contains.
Electrons will not be emitted from the surface of a material if they do not receive sufficient
energy from an incident photon. Since the energy of the incident photon is not dependent on the
intensity, but rather the frequency (and therefore the wavelength), there is a minimum frequency
(the threshold frequency, f0) of incident light required for the photoelectric effect to occur. If the
light is below this threshold frequency, no photoelectrons will be emitted. If the light is above
this threshold frequency, photoelectrons will be emitted. Intensity (the number of photons) does
not matter if the incident light is below the threshold frequency; there will never be any
photoelectrons emitted in this case. If, however, the incident light is above the threshold
frequency, a higher intensity (more photons) will result in more photoelectrons being emitted.
The higher in frequency that a particular photon is above the threshold frequency, the higher the
energy, and therefore the greater the kinetic energy of the emitted photoelectron.
The implication of Einstein’s explanation of the photoelectric effect is that light energy is not
transferred continuously and evenly as the energy of a wave would be, but rather it is transferred
in bundles (quanta, or photons). Einstein’s photon theory was experimentally verified by
Millikan in 1916, and in 1921 Einstein received a Nobel Prize in physics for his photoelectric
effect explanation.
Mathematics of the Photoelectric Effect
Since the energy of light is directly proportional to the frequency of the light, and since a certain
minimum energy of incident light is required to eject electrons from a surface, then there is a
certain minimum frequency of light required to eject electrons from the surface. This minimum
frequency is known as the threshold frequency. There is a corresponding maximum wavelength
of light that will eject electrons from the surface.
The work function is the energy binding an electron to a photoelectric surface, or the minimum
energy required to eject an electron as described above. This work function is equal to Planck’s
constant multiplied by the threshold frequency:
W  hf 0  h
c
max
where W is the work function (J), h is a constant known Planck’s constant (6.63 x 10-34 J.s), f0 is
the threshold frequency of the photoelectric material (Hz), c is the speed of light in a vacuum
(3.00 x 108 m/s), and λmax is the maximum wavelength of light that will eject electrons from the
photoelectric material.
The stopping voltage is that potential difference that would stop the motion of an ejected
photoelectron. In order to stop the motion of an ejected photoelectron, the kinetic energy of the
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ejected photoelectron must be overcome by the electrical energy. This electrical energy is
supplied by the potential difference known as the stopping voltage.
E kmax 
1 2
mvmax  qVstop
2
where m is the mass of an ejected photoelectron (9.11 x 10-31 kg), v is the speed of the ejected
photoelectron (m/s), q is the charge of the ejected photoelectron (C), and Vstop is the stopping
voltage (V). Note that when using either of these equations for kinetic energy, “Joules” must be
used as the units for energy.
It stands to reason that the maximum possible kinetic energy of an ejected photoelectron (it could
be lower than this but not higher) plus the energy used in ejecting the photoelectron (the work
function) should equal the total energy supplied by the incident photon. Einstein derived an
equation to show just that:
hf  Ekmax  W
where h is a constant known as Planck’s constant (6.63 x 10-34 J.s), f is the frequency of the
incident photon (Hz), Ekmax is the maximum kinetic energy of the ejected photoelectron (J), and
W is the work function of the photoelectric material (J).
The above equation is really just another way to state the law of Conservation of Energy!
The photoelectric effect can be demonstrated with the following experimental set-up. The set-up
is similar to one used by Millikan to verify Eintein’s theory.
(Pearson Physics, Ackroyd et al, 2009)
When the power supply is connected as shown and the incoming photons exceed the threshold
frequency (and therefore the work function), a current will be measured by the ammeter because
the photoelectrons will be attracted to the positive collector. If the polarity of the power supply
were reversed, the collector would become negative. It would then repel the photoelectrons and
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the reading on the ammeter would be 0 Amps. The potential difference required to stop the
electrons is the stopping potential, and it can be multiplied by the charge of an electron to get the
maximum kinetic energy.
The following graph can then be plotted:
Maximum Kinetic Energy (x
10-19 J)
Maximum Kinetic Energy vs. Frequency
20
15
10
5
0
-5 0
2
4
6
8
10
12
14
-10
-15
frequency (x 1014 Hz)
The slope of the above graph is Planck’s constant, the x-intercept is the threshold frequency, and
the magnitude of the y-intercept is the work function.
Example 14.4 p.716
Practice Problems p.716
Example 14.5 p.718
Practice Problems p.718
Example 14.6 p.719
Practice Problems p.719
Check and Reflect p.720 #1-10
MC #114-124
Wave-Particle Duality
Diffraction, interference, refraction, and polarization all support the wave theory of light. If,
however, light was wavelike, it would be expected that the intensity, or brightness, of incident
light would determine when photoelectrons were emitted. It is not intensity, however, that
determines this; rather it is the frequency (and therefore according to Einstein, the energy) of the
incident light. Einstein’s notion of bundles, or photons, of light supports a particle theory of
light. We now have the notion of a wave-particle duality of light because light sometimes
behaves like a wave and sometimes behaves like a particle.
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The Compton Effect
In 1923, Arthur Compton noticed that an x-ray photon striking a metallic surface emitted an
electron (as in the photoelectric effect) and a second, less energetic, photon. Classical theory
predicted that the frequency (and therefore energy) of a scattered photon should remain the same
as it was before. In an effort to resolve the conflict, Compton proposed that photons have both
energy and momentum, and that a photon striking a surface was really a collision in which
momentum and energy would be conserved. Since the emitted electron received some of the
initial energy and momentum of the incident photon, the scattered photon would have less
energy and momentum than the incident photon.
The scattering of an x-ray photon by an electron came to be known as Compton scattering, while
the change in energy (and therefore frequency and wavelength) came to be known as the
Compton Effect. The Compton Effect and photon momentum provide evidence to support the
particle nature of light.
The Math:
Since photons have no rest mass, the momentum of a photon cannot be calculated directly using


p  mv . Einstein’s energy-mass equivalence theorem must instead be used:
E
Since mass is equivalent to energy, the equivalent mass of a photon can be expressed as m  2 .
c


 E
 E
 hf
 h
So p  mv becomes p  2 c which becomes p  which becomes p 
and p  .
c
c

c
Compton also derived a relationship between the change in wavelength in the scattered photon
and the direction in which the scattered photon travels. It is based on the laws of conservation of
energy and momentum, and is as follows:
   f  i 
h
(1  cos  )
mc
where Δλ is the change in wavelength in the scattered photon (m), λf is the wavelength of the
scattered photon (m), λf is the wavelength of the incident photon (m), h is Planck’s constant, m is
the mass of the scattering electron (kg), c is the speed of light (3.00 x 108 m/s, and θ is the angle
through which the x-ray scatters.
Example 14.8 p.724
Practice Problems p.724
Example 14.7 p.723
Practice Problems p.723
Check and Reflect p.725 #1-8
MC #125-130
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de Broglie Wavelength
Since the photoelectric effect and Compton scattering demonstrated the particle nature of light,
Louis de Broglie proposed that particles also had a wave nature. He derived an equation for the
 h
“wavelength” of a particle. Since the momentum of a photon was p  , it stood to reason that



the wavelength of a particle could be found be substituting mv for p , and rearranging to find λ.
The de Broglie wavelength of a particle is thus found by using the following equation:
h
p
 
h

mv
where λ is the wavelength of the particle (m), h is a constant known as Planck’s constant (6.63 x

10-34 J.s), m is the mass of the particle (kg), and v is the velocity of the particle (m/s).
Note that the above equation does not appear on the Alberta physics 30 equation sheet and is
not required.
Evidence to support de Broglie’s wave nature of particles hypothesis later came in the form of
electron diffraction experiments. Diffraction is a property of waves, and it turns out that
electrons undergo diffraction.
Then, Now, and Future: The Electron Microscope p.727
MC #131
Unit Review MC #132-145
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Diploma Review Questions
97. Electromagnetic radiation is produced by charged particles that are moving
A. at the speed of light
B. with zero acceleration
C. with a changing velocity
D. parallel to a fixed magnetic field
98. Gamma radiation can be produced by
A. radioactive decay
B. incandescent solids
C. moving charges in a conductor
D. the acceleration of electrons in a television picture tube
99. Regions of the electromagnetic spectrum listed in order from largest to smallest wavelength are
A. X-ray, ultraviolet, visible, infrared, radio
B. X-ray, infrared, visible, ultraviolet, radio
C. radio, ultraviolet, visible, infrared, X-ray
D. radio, infrared, visible, ultraviolet, X-ray
Use the following information to answer the next question.
100. When the regions of the electromagnetic spectrum listed above are arranged order of increasing wavelength, this order is
A. III, I, V, II, IV
B. II, I, V, IV, III
C. III, IV, V, I, II
D. IV, V, III, I, II
-----------------------------------------------------------------101. X-rays are produced by
A. an alternating current of about 1018 Hz
B. firing gamma rays at a tungsten electrode
C. varying the speed of electrons in a magnetic field
D. collisions between high-speed electrons and a metal target
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Use the following information to answer the next question.
102. Match each of the sources of electromagnetic radiation with the type of electromagnetic radiation it produces given below. Use each number
only once.
(Record all four digits of your answer in the numerical-response section on the answer sheet.)
-----------------------------------------------------------------103. Which of the following types of radiation has the longest period?
A. Radio waves
B. Infrared light
C. Ultraviolet light
D. Gamma radiation
Use the following information to answer the next question.
104. The wire mesh layer is necessary because the
A. cable needs a rigid reinforcing layer
B. electric force inside a conductor is not zero
C. electrical signals need to be shielded from strong magnetic and electric fields
D. electrical signals will travel better if they have two different transmitting wires
------------------------------------------------------------------
105. Maxwell’s work contained the new idea that
A. an electric current in a wire produces a magnetic field that circles the wire
B. a current is induced in a conductor that moves across a magnetic field
C. an electric field that changes with time generates a magnetic field
D. two parallel, current-carrying wires exert a force on each other
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Photon Theory
106. The Advanced Composition Explorer (ACE) telescope began operation in August 1997. It detects electromagnetic radiation in the range of
1.0 x 102 eV to 5.0 x 102 MeV. The wavelength range measured by this telescope is
A. 2.0 x 10–27 m to 4.0 x 10–34 m
B. 8.0 x 10–11 m to 1.6 x 10–17 m
C. 1.2 x 10–8 m to 2.5 x 10–15 m
D. 1.2 x 1023 m to 2.4 x 1016 m
Use the following information to answer the next two questions.
107. The energy difference between a laser photon and an emitted photon is
A. 2.00 10–19 J
B. 1.97 10–19 J
C. 2.58 10–21 J
D. 8.62 10–33 J
108. Visible light has frequencies that range between 4.3 1014 Hz (red) and 7.5 1014 Hz (violet). Which of the following statements best
describes the absorbed laser photon and the emitted photon in the optical cooling experiment?
A. Both photons are in the infrared range.
B. Both photons are in the ultraviolet range.
C. Both photons are in the visible light range.
D. One photon is in the visible light range, and one is not in the visible light range.
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Use the following information to answer the next question.
109. Green light with a wavelength of 545 nm reaches the observer’s eyes. The energy of a photon of this green light is __________ eV.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
-----------------------------------------------------------------Use the following information to answer the next question.
110. The mercury atoms emit electromagnetic radiation with a wavelength of 254 nm. The minimum amount of energy that must be transferred
to a mercury atom during excitation to enable this emission, expressed in scientific notation, is b 10–w J. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
-----------------------------------------------------------------Use the following information to answer the next question.
111. A photon of gamma radiation emitted by the radioactive decay of technetium-99 has an energy of 3.85 MeV. This radiation has a
wavelength of
A. 5.17 10–26 m
B. 3.23 10–13 m
C. 3.10 1012 m
D. 9.29 1020 m
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Use the following information to answer the next question.
112. The difference in energy associated with the photons from the two lines of the helium spectrum is
A. 1.60 10–19 J
B. 1.73 10–19 J
C. 4.07 10–19 J
D. 8.14 10–19 J
-----------------------------------------------------------------113. The Compton Gamma Ray Observatory is a satellite that is able to detect electromagnetic radiation from throughout the universe. The
Compton Observatory can detect photons ranging from 4.00 104 eV to 3.00 1010 eV. The highest frequency that can be detected, expressed in
scientific notation, is b 10w Hz. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
Photoelectric Effect
114. A light source with a wavelength of 548 nm shines on a photocell with a 1.60 eV work function. In order to have an output voltage of 12.0
V DC, the number of photocells that must be linked in series is
A. 5 photocells
B. 8 photocells
C. 10 photocells
D. 18 photocells
Use the following information to answer the next question.
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115. What is the maximum kinetic energy of the emitted photoelectrons?
A. 4.91 x 10–19 J
B. 2.91 x 10–19 J
C. 2.00 x 10–19 J
D. 1.28 x 10–19 J
-----------------------------------------------------------------116. The work function of a metal with a threshold frequency of 1.1 1015 Hz, expressed in scientific notation, is a.b 10–cd J. The values of a,
b, c, and d are _____ , _____ , _____ , and _____ .
(Record your four-digit answer in the numerical-response section on the answer sheet.)
Use the following information to answer the next two questions.
117. When light falls on the device at position 1 in the diagram,
A. the Compton effect occurs
B. the photoelectric effect occurs
C. light refraction and diffraction occurs
D. light diffraction and interference occurs
118. Night vision devices have a built-in brightness protection circuit to protect both the device and the viewer from unexpected bright light. The
circuit is activated when the
A. photoelectric current increases
B. photoelectric current decreases
C. kinetic energy of photoelectrons increases
D. kinetic energy of photoelectrons decreases
-----------------------------------------------------------------119. The threshold frequency of light for the emission of photoelectrons from a metal is 4.4 1014 Hz. If light of frequency 6.6 1014 Hz shines
on the metal, then the maximum kinetic energy of the emitted photoelectrons is
A. 7.3 10–19 J
B. 4.4 10–19 J
C. 2.9 10–19 J
D. 1.5 10–19 J
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Use the following information to answer the next three questions.
120. The region of the sound track that will allow the most electrical current to be produced in the phototube is labeled
A. I
B. II
C. III
D. IV
121. The energy that is required to remove the electron from the photoelectric surface in the phototube is called the
A. work function
B. threshold frequency
C. electric potential energy
D. maximum kinetic energy
122. In one second, 1.45 1016 photons are incident on the phototube. If each of the photons has a frequency greater than the threshold
frequency, then the maximum current to the amplifier, expressed in scientific notation, is a.bc 10–d A. The values of a, b, c, and d are _____ ,
_____ , _____ , and _____ .
(Record your four-digit answer in the numerical-response section on the answer sheet.)
-----------------------------------------------------------------123. In the photoelectric equation, the symbol W represents the
A. energy gain of the target metal
B. wavelength of the incident radiation
C. maximum wavelength of an emitted electron
D. minimum energy required to release an electron from a metal
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124. Violet light striking the negative electrode in a phototube causes a current to flow in the tube. Under the same conditions, another form of
light that will always cause a current to flow is
A. blue
B. green
C. infrared
D. ultraviolet
Compton Effect
Use the following information to answer the next two questions.
125. The name given to interaction I is
A. Lenz’s Law
B. X-ray production
C. the Compton effect
D. the de Broglie hypothesis
126. The reason that pair production occurs, rather than the production of a single electron, is that the production of a single electron would
violate the Law of Conservation of
A. Mass
B. Charge
C. Energy
D. Momentum
-----------------------------------------------------------------127. A photon exhibits properties of a particle because it has
A. mass
B. momentum
C. a constant speed
D. a fixed frequency
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Use the following information to answer the next question.
128. A photon has a momentum of 4.0 10–23 N.s. The frequency of the photon, expressed in scientific notation, is a.b 10cd Hz. The values of
a, b, c, and d are _____ , _____ , _____ , and _____ .
(Record your four-digit answer in the numerical-response section on the answer sheet.)
-----------------------------------------------------------------129. A result that emerged from Einstein’s work is the expression E = pc, where p is the magnitude of the momentum of a photon. The
magnitude of the momentum of a 1.30 102 eV photon is
A. 6.93 10–26 kg.m/s
B. 2.08 10–17 kg.m/s
C. 8.20 10–14 kg.m/s
D. 4.33 10–7 kg.m/s
130. The particle nature of X-ray radiation is best demonstrated by the observation that X-rays
A. exhibit the Compton effect
B. have great penetrating ability
C. are diffracted by pure crystals
D. are not deflected by magnetic fields
de Broglie Waves
Use the following information to answer the next question.
131. In order for a virus to be detected by an electron microscope, the minimum speed that the electrons must have in the electron microscope,
expressed in scientific notation, is b 10w m/s. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
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Electromagnetic Radiation Review Questions
Use the following information to answer the next two questions.
132. The amount of time, in days, that it takes the radio waves detected by the telescope to reach Earth is
A. 2.7 × 108 days
B. 6.5 × 109 days
C. 2.3 × 1013 days
D. 2.0 × 1018 days
133. DRAO is located in a basin surrounded by mountains, which shield it from manmade radio waves that interfere with astronomical signals.
Manmade radio waves are produced by
A. radioactive decay
B. electron transitions in atoms
C. oscillating charges in a linear antenna
D. high speed electrons stopped suddenly by a metal surface
Use the following information to answer the next two questions.
134. Given the information above, the element that emits the lowest energy photon of visible light is
A. strontium
B. barium
C. copper
D. sodium
135. The colours are emitted by electrons that are
A. undergoing transitions to higher energy levels
B. undergoing transitions to lower energy levels
C. oscillating between energy levels
D. emitted by the nucleus
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136. A metal has a work function of 2.91 × 10 –19 J. Light with a frequency of 8.26 × 1014 Hz is incident on the metal. The stopping voltage is
__________V.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
137. If a light with a wavelength of 3.25 × 10–8 m illuminates a metal surface with work function of 5.60 × 10 –19 J, the maximum kinetic energy
of the emitted photoelectrons is
A. 5.60 × 10–19 J
B. 5.56 × 10–18 J
C. 6.12 × 10–18 J
D. 6.68 × 10–18 J
Use the following information to answer the next three questions.
138. The type of light indicated by point P is
A. visible
B. infrared
C. microwave
D. ultraviolet
139. The energy of a photon of light indicated by point P is
A. 4.1 eV
B. 2.3 eV
C. 1.7 eV
D. 0.0 eV
140. Photons of light, as indicated by point P, bombard the cesium plate. The maximum kinetic energy of an emitted electron is
A. 4.1 eV
B. 2.3 eV
C. 1.7 eV
D. 0.0 eV
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141. The Compton experiment was significant in that it demonstrated that photons have
A. mass
B. momentum
C. wave properties
D. a speed of 3.00 × 108 m/s
Use the following information to answer the next question.
142. The concept that explains the collapse of the rabbit fur is
A. induction
B. grounding
C. conduction
D. the photoelectric effect
------------------------------------------------------------------
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Use the following information to answer the next question.
143. The concept that explains the collapse of the rabbit fur is
A. induction
B. grounding
C. conduction
D. the photoelectric effect
-----------------------------------------------------------------Use the following information to answer the next question.
144. Based on the graph above, Planck’s constant has a value of
A. 6.6 10–34 J.s
B. 5.0 10–34 J.s
C. 3.6 10–34 J.s
D. 3.0 10–34 J.s
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Use the following information to answer the next question.
145. The work function of the material emitting the photoelectrons is
A. 2.0 1015 J
B. 1.3 10–18 J
C. 6.6 10–34 J
D. 0.0 J
------------------------------------------------------------------
EMR MC Answers
97
98
99
100
101
102
103
104
105
106
107
108
109
110
C
A
D
C
D
3124
A
C
C
C
C
A
2.28
7.83
B. Dickie
111
112
113
114
115
116
117
118
119
120
B
B
7.25
D
C
7319
B
A
D
C
121
122
123
124
125
126
127
128
129
130
A
2323
D
D
C
B
B
1819
A
A
131
132
133
134
135
136
137
138
139
140
1.45
A
C
A
B
1.60
B
D
A
B
141
142
143
144
145
B
D
D
B
B
117
Cathode Ray Tubes
In 1855, Heinrich Geissler developed a powerful vacuum pump. Shortly thereafter, Julius
Plücker sealed a wire into either end of a Geissler tube, attaching the wires to a battery outside of
the tube. Plücker found that electricity passed through the low-pressure gas in the tube, and that
a greenish glow filled the tube. This was an early version of the cathode ray tube.
William Crookes later did experiments which led him to believe that the greenish glow observed
in the Geissler tube consisted of particles.
In 1897 J.J. Thomson performed experiments on cathode rays that led to him being credited with
the discovery of the electron.
Thomson’s Cathode Ray Tube
1. acceleration chamber
2. velocity selection chamber
3. main chamber
Thomson’s cathode ray tube consisted of three main chambers:
3. Main Chamber
It was well known before Thomson began his experiments that cathode rays could be deflected
by electric and magnetic fields. From these deflections, Thomson reasoned that the cathode rays
were really beams of negatively charged particles (electrons). He further reasoned that he could
measure the charge to mass ratio of these electrons by passing a cathode ray through a magnetic
field and measuring the radius of its path (the path of charged particles in a uniform magnetic
field is circular). The centripetal force experienced by the cathode ray would be caused by the
magnetic force. Therefore,
B. Dickie
118
Fm  Fc
v2
r
v
qB  m
r
q
v

m Br
qvB  m
where q/m is the charge to mass ratio (C/kg), v is the speed of the cathode ray (m/s), B is the
strength of the magnetic field (T), and r is the radius of curvature of the cathode ray (m).
2. Velocity Selection Chamber
To find the speed of the cathode ray to substitute into the above equation, one could use a
velocity selection chamber in the cathode ray tube. In a velocity selection chamber, a magnetic
field and electric field are adjusted until the magnetic force and electric force are balanced for all
electrons moving at a particular speed. These electrons will travel undeflected through the
velocity selection chamber and enter the main chamber. Electrons moving at other speeds will
be deflected and not enter the main chamber.
Fe  Fm

q E  qvB

E  vB

E
v
B

where v is the speed of the cathode ray (m/s), E is the electric field strength (N/C), and B is the
magnetic field strength (T).
1. Acceleration Chamber
The cathode rays (electrons) are first accelerated across a potential difference using electrical
potential energy. The electrical potential energy is converted to kinetic energy and the electrons
gain speed. Thus, conservation of energy can be used to analyze the acceleration chamber:
E A  EB
qV 
1 2
mv
2
where q is the charge of an electron (1.60 x 10-19 C), V is the potential difference in the chamber
(V), m is the mass of an electron (9.11 x 10-31 kg), and v is the speed at which the electrons exit
the acceleration chamber (m/s).
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The significance of Thomson’s experiments was that there became a recognition that the atom
contained negatively charged particles called electrons.
Example 15.1 p.756
Practice Problems p.756
Example 15.2 p.758
Practice Problems p.758
Then, Now, and Future: The Mass Spectrometer p.759
Check and Reflect p.760 #1-10
MC #146-149
Thomson’s Atomic Model
Thomson performed a series of experiments that led him to conclude that the greenish cathode
rays were really beams of negatively charged particles (electrons). Thomson thus developed an
atomic model that was different from the previous Dalton Solid Sphere model. Instead of the
atom being a simple solid sphere, Thomson said that the atom was a sea of positive charge filled
with these negatively charged electrons.
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Millikan’s Oil Drop Experiment
Between 1909 and 1916, Robert Millikan designed an experiment that would measure the
elementary charge (the charge on one electron). Millikan’s experiment is called the “Millikan
Oil Drop Experiment”.
Millikan sprayed oil through an atomizer (kind of like a perfume bottle dispenser) into a chamber
that was bordered on the top by a positively charged plate and on the bottom by a negatively
charged plate. The oil droplets acquired either a positive or negative charge due to friction as
they passed through the atomizer. The positively charged oil drops accelerated downward due to
the influence of gravity and the electric field between the plates. The negatively charged oil
drops would accelerate upward if the upward electric force was larger than the gravitational
force, but would be suspended or travel up or down at a constant velocity If the two forces were
equal.
The easiest oil drops to analyze are those that are suspended or are traveling at a constant
velocity:
Millikan’s Oil Drop Experiment Set-Up
Fe  Fg

q E  mg
V
 mg
d
mgd
q
V
q
Through several trials, Millikan was able to find the charge on many different oil drops. He
noticed that all of the charges of the oil drops were multiples of 1.60 x 10-19 C. He therefore
deduced that all other charges would be multiples of this charge. Thus, charge is said to be
quantized.
So how should you analyze an oil droplet that is not suspended or traveling at a constant
velocity? If the droplet is accelerating either up or down, we use the following analysis:
B. Dickie
121

 
Fnet  Fe  Fg


ma  q E  mg
V

ma  q  mg
d

Note that a is positive if the droplet is accelerating upward and negative if it is accelerating

downward. Fe will be positive or negative depending on the direction of the electric force, and

Fg will always be negative. Remember to not plug in the signs until you start plugging in
numbers and that we always ignore the sign of the charge when plugging the value of the charge
into equations.
Example 15.3 p.763
Practice Problems p.763
Example 15.4 p.764
Practice Problems p.764
Check and Reflect p.767 #1-7
MC #150
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122
Isotope Notation
The nucleus of an atom is made up of two constituents, or nucleons. The two types of nucleons
are protons and neutrons.
Nucleon
proton
neutron
Charge
1.60 x 10-19 C
0C
Mass
1.6726 x 10-27 kg
1.6749 x 10-27 kg
The number of protons in the nucleus of an atom is known as the atomic number.
designated by the letter Z.
It is
The total number of nucleons (protons + neutrons) is called the mass number. It is designated by
the letter A. To find the number of neutrons, one can simply subtract the atomic number from
the mass number.
To specify a particular nucleus, we can use isotope notation:
A
Z
X
where X is the chemical symbol for the element.
Note that the atomic number for each element appears on the periodic table, however the mass
number does not.
Example 16.1 p.790
Practice Problems p.790
MC #151
B. Dickie
123
Radioactivity
Henri Becquerel accidentally discovered radioactivity in 1896, two months after the discovery of
x-rays. While attempting to determine the connection between x-rays and fluorescence or
phosphorescence, he found that a certain mineral darkened a photographic plate even when the
plate was covered from light. Unlike x-rays, this mineral emitted radiation without external
stimulus. This phenomenon came to be known as radioactivity. Becquerel also discovered that
this radiation could ionize molecules, leading to the term ionizing radiation.
Becquerel’s colleagues, Marie and Pierre Curie, soon discovered other radioactive elements,
proving that radioactivity was not just a property of one element.
In 1899, Ernest Rutherford discovered two specific types of radioactive decay, alpha (α) and
beta (β) decay. In 1900, Paul Villard discovered gamma decay.
Particle
Symbol

Particle /
Radiation
particle (helium
nucleus)

particle
(electron)
alpha
4
2
beta
0
1
gamma

0
0
high
energy
electromagnetic
radiation
Penetrating Power
large size makes
penetrating
power
very short (a piece of
paper will stop an
alpha particle)
smaller size allows
them to penetrate 1-2
cm of human flesh
very high; can pass
straight through the
human body and
many centimetres of
lead
Ionizing Power
very high
moderate
low
In beta decay, an antineutrino (a neutral particle with a very small, but unknown, mass) is also
emitted. In 1930, Wolfgang Pauli theorized the existence of antineutrions to account for missing
kinetic energy when beta decay occurred. The symbol for an antineutrino is .
Note that a second type of beta decay called positron emission can also occur. A positron ( 10  )
has the same properties as those of a beta particle, except that it is positively charged. We
therefore say it is the antiparticle to a beta particle. When positron emission occurs, a neutrino
is also produced. The symbol for a neutrino is .
B. Dickie
124
Absorption of Ionizing Radiation and its Effects
All ionizing radiation (alpha, beta, and gamma) has the ability to energize electrons in the atom
to the extent that they leave the atom (i.e. the atom becomes ionized). This can cause direct
damage to DNA or RNA. It can also result in normally unreactive atoms becoming charged
atoms (ions) that are quite reactive. Undesirable chemical reactions can take place between these
ions that, in humans, can cause mutations in genetic material, vomiting, cataracts in the eyes,
cancer, and a weakening of the immune system.
In addition, molecular damage can cause a disruption to the functions taking place within cells,
leading to the possibility of death of the cells or the prevention of the cells from multiplying.
Three types of ionizing radiation vary in their degrees of penetrating power and ionizing power,
and can therefore produce varying effects.
Alpha radiation is generally not serious because of its low penetrating power. If it is ingested,
however, it can be quite serious because of its high ionizing power.
Beta radiation has a higher penetrating power (1-2 cm of human flesh) but a lower ionizing
power.
Gamma radiation has the lowest ionizing power. This can be offset, however, by gamma
radiation’s very high penetrating power.
The seriousness of exposure also depends, of course, on the rate of decay of the radioactive
nucleus (the higher the rate of decay, the higher the exposure) and the intensity of the radiation
(the higher the intensity, the higher the exposure). Obviously the lowest possible exposure is
desirable.
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125
Nuclear Stability
There are two main forces that act within the nucleus: the electrostatic force and the strong
nuclear force.
The electrostatic force is a repulsive force that causes all of the protons in the nucleus to repel
each other. This would cause the nucleus to fly apart if it were not for the strong nuclear force.
The strong nuclear force is an attractive force between protons and neutrons that is strong, but
short-ranged. Small nuclei that are stable tend to have roughly the same number of protons as
neutrons, while larger stable nuclei tend to have proportionately more neutrons. The repulsive
force between the protons in small nuclei is not big enough to cause the nucleus to break apart
because of the small number of protons, however the repulsive force between the protons in large
nuclei would be large enough to cause the nucleus to break apart if not for a larger number of
neutrons causing a larger strong nuclear attractive force, keeping the nucleus together.
(Pearson Physics, Ackroyd et al, 2009)
The above graph is called a Segre plot and it shows the stability of nuclides. There are roughly
2500 different nuclides, but only about 300 of them are stable. The dotted line on the above plot
shows nuclei that have the same number of protons as neutrons. The black dots show the band
of stable isotopes. Note that at small atomic numbers, stable isotopes have similar numbers of
protons and neutrons. However with larger nuclei, stable nuclei have more neutrons than
protons. No nuclei above atomic number 82, however, will be stable because there are simply
too many protons. Nuclei above the band (nuclei with too many neutrons) will be unstable
because the nucleus will energetically favour decays of neutrons into protons and beta particles
B. Dickie
126
(beta decay). This beta decay can happen when there are too many neutrons in a small or large
nucleus. Note that an antineutrino is also produced. Nuclei below the band will undergo
positron emission, where a proton decays into a neutron and a positron (and a neutrino).
Gamma decay occurs when the nucleus is left in an excited state. Nuclei, like orbital electrons,
can only exist in certain energy states. When the nucleus is left in an excited state (as a result of
alpha decay, beta decay, or some other process), it makes a transition back to a lower state
emitting a photon of gamma radiation.
MC #152-154
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127
Nuclear Decay and Transmutation
In the process of alpha and beta decay, the nucleus of one element is transformed into the
nucleus of another. This process is called transmutation.
It is important to note that in any transmutation equation, the total of the mass numbers (the
superscript “A”) on the left side of the equation must be equal to the total of the mass numbers
on the right side. In addition, the total of the atomic numbers (the subscript “Z”) on the left side
must be equal to the total of the atomic numbers on the right side. Problems may involve being
given a radioactive element and being told that it undergoes alpha (or beta) decay. The problem
solving strategy is to write the left side of the equation (the radioactive element) and then the part
of the right side that you know  24 or 10  . Figure out what mass number and atomic number is
missing, then find the element on the periodic table with the corresponding atomic number and
write it down.
An Example Not in Your Book
Write the nuclear equation for the transmutation of a radioactive isotope
decay.
226
88
Ra undergoing alpha
Another Example Not in Your Book
Write the nuclear equation for the transmutation of a radioactive isotope
decay.
214
82
Pb undergoing beta
Yet Another Example Not in Your Book
Complete the following nuclear equations:
a.
30
15
30
P16
S ?
U 234
90Th  ?
b.
238
92
c.
13
7
N 136C  ?
B. Dickie
128
What? There’s More?
What remains after
234
91
Pa undergoes a beta decay, followed by two alpha decays?
Still More Examples Not in Your Book
Complete the following nuclear artificial transmutation equations:
N  24  178O  ?
a.
14
7
b.
105
48
Cd  10  ?
Will This Ever be Over???
235
92
U is transformed into what isotope after two alpha and two beta decays?
Practice Problems p.799
Practice Problems p.800
Practice Problems p.803
Check and Reflect p.810 #1-7, 9, 11
MC #155-158
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129
Half Life
Half-life is defined as the time that it takes one half of a radioactive sample to decay.
One might think that since half of a radioactive substance will decay after one half life, then the
remaining half will decay after the second half life. This is not the case. It is important to note
that half-life is a statistical probability that a given radioactive nucleus will decay. On average,
one-half of remaining radioactive nuclei decay in one half-life. This means that after the second
half-life, only half of what remained from the first half-life has decayed. Radioactivity is an
exponential process.
One can calculate how many (or activity of or mass of) radioactive nuclei of a particular element
remain after a given amount of time by using the following equation:
1
N  N0  
2
n
where N is the number of radioactive nuclei remaining (or activity remaining in decays/second or
Becquerel (Bq), or mass remaining in grams or kilograms), N0 is the original number of
radioactive nuclei (or original activity in decays/second or Becquerel (Bq), or original mass in
grams or kilograms), and n is the number of half-lives (note: not the amount of time) that have
elapsed (unitless).
To find the number of half lives n, use the following equation:
n
t
t1
2
where n is the number of half lives, t is the time elapsed (any time units), and t 1 is the half life
2
(same units as t).
Alternatively, n can sometimes be found by dividing the value of N0 by 2 until the value of N is
obtained, then counting how many times you divided. Using logarithms is also an option.
Example 16.13 p.814
Practice Problems p.814
An Example Not in Your Notes:
The half life of carbon-14 is 5730 years. How long does it take 12.5 kg of carbon-14 to decay
into 400 g of carbon-14?
B. Dickie
130
A graph of radioactive decay can take several forms. The y-axis will can be Activity (Bq), Mass
(kg), or % remaining (%). In general, the y-axis will be N. The x-axis is typically time, however
time can be measured in any units. To determine the half-life of a radioactive sample from a
graph, one should first find the point on the y-axis where only half of the original
activity/mass/% remains. Draw a horizontal line straight across until it meets the curve, then
draw a vertical line straight down to the x-axis. The point where the vertical line intersects with
the x-axis is the half-life of the substance.
http://www.bbc.co.uk/schools/gcsebitesize/science/images/12_half_life.gif
On the above graph, 50% of the mass remains after 2.0 days, therefore the half-life of the
substance is 2.0 days.
Check and Reflect p.817 #1,3-8
Then, Now, And Future: Radiotherapy
MC #159-163
B. Dickie
131
Nuclear Fission and Fusion
Nuclear Fission is the process by which heavy nuclei are split into two or more nuclei of
intermediate numbers.
For example,
1
0
92
141
1
n 235
92 U 36 Kr  56 Ba 30 n  200MeV
A slow neutron is fired at uranium-235. This produces uranium-236. Because this is unstable, it
decays into any one of a number of pairs of nuclei (in this case krypton-92 and barium-141).
This releases about 200 MeV of energy and three more neutrons. If one or more of these
neutrons causes another fission reaction, the process is called a chain reaction.
Nuclear Fusion is the joining together of light nuclei into a single, heavier nucleus. In the
process, energy is released. It is the process that powers the Sun.
For example,
2
1
H  12H 23He 01n  energy
The problem that must be overcome with fusion is that the nuclei must be given a lot of energy
to overcome the repulsive electric force (recall nuclei are positively charged). In the H-bomb
(1952), this energy was supplied by an atomic explosion. In any case, the temperature must be
very high.
B. Dickie
132
Mass Defect and Binding Energy
Mass defect can be defined as the difference in mass between the nucleons in a nucleus and the
total mass of the nucleus. The energy equivalent of mass defect is called binding energy.
Nuclear reactions (i.e. fission and fusion) cause nuclei to transmutate into new nuclei that have a
lower total mass than the original nuclei. The energy released in a nuclear reaction corresponds
to the difference in mass, and therefore energy, of the reactants and products. It can be found by
using Einstein’s energy-mass equivalence equation:
E  mc 2
where E is energy (J), m is mass (kg), and c is the speed of light in a vacuum (3.00 x 108 m/s).
Masses are often given to you in atomic mass units (u). One atomic mass unit is 1/12 of the
mass of a carbon-12 atom. One atomic mass unit has a mass of 1.66 x 10-27 kg and an equivalent
energy of 931.5 MeV. Note that any masses given to you in atomic mass units must be
converted to kilograms before using the above equation.
Note also the following two sets of mass units: MeV/c2 and GeV/c2. These both come from the
E
equation m  2 . 1.00 MeV/c2 is equal to 1.78 x 10-30 kg, while 1.00 GeV/c2 is equal to 1.78 x
c
10-27 kg.
Example 16.3 p.794
Practice Problems p.794
Example 16.14 p.819
Practice Problems p.819
MC #164-169
B. Dickie
133
(Pearson Physics, Ackroyd et al, 2009)
B. Dickie
134
Rutherford’s Gold Foil Experiment
Rutherford bombarded gold foil with a stream of alpha particles. Most of the positively charged
alpha particles penetrated the foil easily, as indicated by a flash of light when they struck a zinc
sulfide screen, and a few were slightly deflected. Some, however, were deflected at extreme
angles and some even deflected backwards. This could not be explained by Thomson’s model.
(http://www.britannica.com/ebc/art-18080/The-Rutherford-gold-foil-experiment-In-1911-physicist-ErnestRutherford)
Rutherford’s explanation: Positive charge is concentrated in a very small region of the atom and
the electrons (negative charges) are distributed in the space around this positive charge. If the
positively charged alpha particles were mostly undeflected or deflected very little, most of the
atom must not be positively charged (i.e. no repulsion). A small area must be positively charged
and very heavy relative to the rest of the atom, however, if a few particles are to be scattered.
This was essentially the discovery of the nucleus, as Rutherford was able to develop formulas to
predict observed scattering. Rutherford also said the electrons were circling around the positive
charge (nucleus), held in orbit by electrostatic force (just as planets circle around the Sun and are
held in orbit by gravity) at a radius of approximately 10-15 m. As a result of his observations and
assumptions, Rutherford’s model became known as the planetary model.
Check and Reflect p.770 #1, 2, 4
MC #172
B. Dickie
135
The Bohr Model of the Atom
Bohr’s model of the atom is as follows:
1. Only a few special orbits are allowed and each is characterized by a specific amount of
electron energy.
2. If electrons orbit the nucleus, they must experience a centripetal force. They therefore must
be accelerating toward the centre. Recall that Maxwell said that accelerating charges would
continuously emit energy in the form of electromagnetic waves. If energy were continuously
given off, however, the electron should spiral into the centre (because of reduced speed and
therefore reduced radius). This does not happen, however. The size and shape of the orbit,
along with its energy, remain constant. The orbits are therefore often called stationary states.
3. Electrons may switch energy levels:
a. They can jump to a higher orbit by absorbing energy. A dark line is produced.
Specific energies are associated with specific electron levels with the hydrogen atom. In order for an
electron to jump from a lower level to a higher level, a specific amount of energy corresponding to the
difference in energies between the two levels must be absorbed. Different transitions require different
amounts of energies, and therefore result in a spectrum of energy being absorbed.
b. They can fall from a higher orbit to a lower orbit. In the process they give off energy in
the form of a single photon. A bright line is produced.
A line emission spectrum appears as the reverse of a line absorption spectrum.
B. Dickie
136
Continuous vs. Line Spectra
All substances radiate electromagnetic radiation when they are heated. Since electromagnetic
radiation is emitted when electrons fall to a lower level and since the electron configuration is
different in each element, such emitted radiation can be expected to provide clues about an
element’s atomic structure. These electromagnetic waves are analyzed by a spectrometer, which
uses a prism or a diffraction grating to organize the radiation into a spectrum. ). For a very hot
gas under low pressure, however, the spectrum consists of a series of bright lines (corresponding
to frequencies of light emitted) separated by dark regions. This is called a line emission
spectrum. The line emission spectrum for each element is different and can thus be used to
identify the element.
Of course whenever radiation is absorbed by an element, an electron can rise to a higher level in
the atom. A line absorption spectrum is thus produced. A line absorption spectrum appears as
the opposite of a line emission spectrum. Wherever you would see dark on an emission
spectrum, you would see light (like a rainbow) on an absorption spectrum. Wherever you would
see light (bright lines) on an emission spectrum, you would see dark lines (corresponding to
frequencies of light absorbed) on an absorption spectrum.
For heated solids, liquids, and gases under high pressure, the spectrum is continuous (i.e. it
contains all wavelengths and is similar to a rainbow. The radiation produced in a continuous
spectrum is assumed to be due to oscillations of atoms and molecules.
(http://www.scienceinschool.org/repository/images/issue4spectrometer11_large.jpg)
Some examples of line emission spectra are shown below. Keep in mind that if the diagrams
were in colour, the lines shown would all be different colours corresponding to the wavelengths
given
B. Dickie
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B. Dickie
138
An Example Not in Your Book
Using the line emission spectra above, identify which elements are present in the unknown gas
mixture whose spectrum is shown below.
The wavelength of a spectral line (absorption or emission) can be found with by first finding the
energy of the transition. To do this, simply subtract the energy of the initial level from the
c
energy of the final level. Then use the equation E  h to determine the wavelength of the

light involved in the transition.
B. Dickie
139
Electron Level Diagrams
The following is a diagram representing several series of electron transitions. We call a diagram
like this an electron level diagram.
Energy
0.0 eV
-0.38 eV
-0.55 eV
-0.85 eV
n=
∞
6
5
4
-1.51 eV
3
-3.4 eV
2
-13.6 eV
1
Free electron
Excited state
Ground State
Lyman
Balmer
Paschen Brackett
To find the energy involved in an electron jumping or falling from one level to another using an
electron level diagram, simply take the difference in energies associated with the two levels
concerned. Remember that it is energy emitted if the electron is falling from a higher level to a
lower level, while it is energy absorbed if the electron is jumping from a lower level to a higher
level.
An Example Not in Your Book
Using the diagram above, determine how much energy is emitted in the second line in the
Balmer series in the hydrogen atom?
Another Example Not in Your Book
Using the diagram above, determine the ionization energy for an electron on the first level of the
hydrogen atom.
Example 15.6 p.775
Practice Problems p.775
MC #173-174
B. Dickie
140
The Schrodinger Model of the Atom
The Schrodinger model fundamentally assumes that there is a wave associated to the electrons
orbiting the nucleus. This wave is called the wavefunction. Because of the wave nature of the
electron, the theory gives only probabilities for finding an electron somewhere rather than
pinpointing a particular point in space. The square of the absolute value of the wavefunction can
be used to find the probability that an electron will be found at a certain position at a certain
time. The wave function collapses when it an observation is made, and the electron will then be
observed to be in one location.
A Summary of Some Atomic Models
Model
Positively
charged
nucleus?
Yes
Electrons
revolve around
nucleus?
Yes
Electron levels?
No
Fixed radius
for electron
levels?
-----
Bohr
Yes
Yes
Yes
Yes
Quantum
mechanical
Yes
Yes
Yes
No
Rutherford
B. Dickie
141
Detecting Subatomic Particles
Physicists now know that two of the three subatomic particles that we normally consider to be
fundamental are not fundamental at all. While electrons are still considered to be fundamental,
protons and neutrons are no known to be made up of smaller particles called quarks. There are
many other fundamental particles besides quarks and electrons, but that will be left for another
time!
Scientists have measured the properties of subatomic particles by accelerating particles such as
protons to extremely high speeds, colliding them with other particles to break them apart, and
tracking the paths of what is produced. Extremely high energies are needed in these collisions to
overcome the strong nuclear force within the nucleus and the weak nuclear force within the
subatomic particles themselves.
There are several ways to track the particles produced after a collision has occurred, including
using cloud chambers and bubble chambers:
1. A cloud chamber contains dust-free air supersaturated with water or alcohol. A charged
particle moving through the air in the chamber can ionize some molecules along its path
and can cause vapour to condense into droplets of liquid (i.e. a cloud) along the path of
the particle, thus showing the path of the particle.
2. A bubble chamber contains a liquefied gas at a low pressure. This results in a depressed
boiling point of the liquid. When the boiling point is reduced to a value just below the
actual temperature of the liquid, ions formed by moving charged particles will cause the
liquid to boil, leaving a track showing the particle’s path.
In both of the above, a magnetic field is applied across the chamber causing the direction of the
moving charged particles to change. The resulting tracks can be analyzed to determine the mass
and charge of the particles. If a single track happens to branch into two or more tracks, it
indicates that the original particle has split up into two or more different particles.
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The Particles
Scientific investigation has shown that there are two families of particles: leptons and hadrons:
1. Leptons are particles such as electrons and electron-neutrinos (and their antiparticles
positrons and antineutrinos) that do not interact via the strong nuclear force. In total,
there are six known types of leptons and their six corresponding anti-particles.
Particle
electron
electron-neutrino
Symbol
eυe
Mass-Energy (MeV/c2)
0.511
<7 x 10-6
Lifetime
Stable
unknown
2. Hadrons are particles such as protons and neutrons that do interact via the strong nuclear
force. Hadrons are made of smaller particles called quarks. There are three generations
of quarks containing a total of six different types of quarks, but protons and neutrons are
both composed of only the first generation quarks (the up quark and the down quark).
Particle
up quark
down quark
Symbol
u
d
Mass-Energy (MeV/c2)
1.5-4
4-8
Charge
+2/3 e
-1/3 e
A proton consists of a down quark and two up quarks, and a neutron consists an up quark
and two down quarks.
Subatomic Particles and Beta Decay
As you know from a previous section, beta-negative decay occurs when a neutron is transformed
into a proton, an electron, and an antineutrino. The decay can be written as follows:
Beta-plus decay occurs when a proton turns into a down quark by emitting a virtual particle W+
that then decays into a positron and a neutrino. The decay can be written as follows:
Notice that charge is conserved in each of the above.
Unit Review MC #175-198
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Diploma Exam Review Questions
Cathode Ray Tube / Mass Spectrometer
Use the following information to answer the next question.
146. An electron hits the screen at a speed of
A. 1.0 107 m/s
B. 1.5 107 m/s
C. 3.0 107 m/s
D. 8.8 1014 m/s
-----------------------------------------------------------------147. J. J. Thomson’s experiments indicated that cathode rays are
A. photons
B. electromagnetic radiation
C. positively charged particles
D. negatively charged particles
Use the following information to answer the next two questions.
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148. The particles in the undeflected beam are moving at a speed of a.b x 10c m/s. The values of a, b, and c, are, respectively, ____, ____, and
____ .
(Record your three-digit answer in the numerical-response section on the answer sheet.)
149. Using the charge-to-mass ratio of the particles, the scientist determines the particles to be
A. protons
B. neutrons
C. electrons
D. alpha particles
-----------------------------------------------------------------Millikan Oil Drop Experiment
150. Two scientists who conducted experiments that led to the determination of the mass of an electron were
A. Planck and Einstein
B. Rutherford and Bohr
C. Thomson and Millikan
D. Compton and de Broglie
Isotope Notation
218
151. Polonium has more isotopes than any other element, and they are all radioactive. The isotope 84 Po
A. 218 protons and 84 neutrons
B. 84 protons and 218 neutrons
C. 134 protons and 84 neutrons
D. 84 protons and 134 neutrons
Radioactivity (Qualitative)
152. Nuclear radiation exists in several different forms. Listed from greatest to least in their ability to penetrate human tissue, the order of three
of these forms is
A. alpha, beta, gamma
B. gamma, beta, alpha
C. gamma, alpha, beta
D. alpha, gamma, beta
153. When a neutral meson particle °) decays, it produces an electron (e – ). In this process, it is most likely that
A. nothing else is produced
B. a gamma ray is also produced
C. a negative particle is also produced
D. a positive particle is also produced
Use the following information to answer the next question.
154. Which of the following situations would result in a person experiencing the most exposure to radioactivity?
A. Being exposed to isotope X at a distance of two metres for two hours
B. Being exposed to isotope X at a distance of one metre for two hours
C. Being exposed to isotope Y at a distance of two metres for two hours
D. Being exposed to isotope Y at a distance of one metre for two hours
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Transmutation Equations
Use the following information to answer the next question.
155. In the transmutation reaction above, an alpha particle is absorbed by a nitrogen nucleus. An unstable nucleus that decays by producing a
A
proton and an unidentified nucleus Z
X
is produced. The values of A and Z are, respectively,
A. 16 and 9
B. 15 and 8
C. 11 and 6
D. 8 and 15
Use the following information to answer the next question.
156. The product of the carbon-14 decay is
14
A. 7
N
14
B. 8 O
10
C. 4 Be
12
D. 6 C
-----------------------------------------------------------------Use the following information to answer the next question.
157. If the air is ionized by alpha particles produced by the americium-243, what immediate byproduct would one expect to find?
A. Curium-243
B. Plutonium-243
C. Berkelium-247
D. Neptunium-239
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Use the following information to answer the next question.
158. In both decays I and II, the type of emitted particle is
A. an alpha particle
B. an electron
C. a neutron
D. a proton
-----------------------------------------------------------------Half-Life Problems
Use the following information to answer the next question.
159. The mass of 224 Ra remaining after 22 days is
A. 0.16 g
B. 0.31 g
C. 2.7 g
D. 3.7 g
-----------------------------------------------------------------Use the following information to answer the next question.
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160. An archeological sample is dated using the carbon-14 dating process and is found to be 2 865 years old. What percentage of the original
carbon-14 remains?
A. 25.0%
B. 29.3%
C. 70.7%
D. 75.0%
-----------------------------------------------------------------161. For a 768 g sample of an unknown radioactive element, 48.0 g remain after 10.2 h. The half-life of the element is __________ h.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
Use the following information to answer the next question.
162. If the biological processes that might eliminate some of the technetium-99 from the body are ignored, the maximum percentage of
radioactive technetium-99 that could still be present in a patient’s system 24.0 h after injection is
A. 12.5%
B. 6.25%
C. 2.00%
D. 0.841%
-----------------------------------------------------------------Use the following information to answer the next question.
163. Americium-243 has a half-life of approximately 7 000 years. If a detector containing 20 mg of this isotope were discarded and then
rediscovered 70 years later, approximately how much americium-243 would remain?
A. 20 mg
B. 0.20 mg
C. 2.0 10–7 mg
D. No measurable amount would remain.
------------------------------------------------------------------
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Fission / Fusion
Use the following information to answer the next two questions.
164. The value of cd in the above reaction can be identified using the Law of Conservation of
A. Mass
B. Energy
C. Charge
D. Momentum
ab
165. The fission product in this reaction is represented by cd
X
. The values of a, b, c, and d are _____ , _____ , _____ , and _____ .
(Record your four-digit answer in the numerical-response section on the answer sheet.)
-----------------------------------------------------------------Use the following information to answer the next three questions.
166. The value of x in reaction II is
A. 4
B. 3
C. 2
D. 1
Use the following additional information to answer the next two questions.
167. The mass defect of uranium-235, expressed in scientific notation, is b 10–w kg. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
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168. The nuclear binding energy of uranium-235, expressed in scientific notation, is b 10w eV. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
-----------------------------------------------------------------Use the following information to answer the next question.
169. In the above fission reaction, the mass of the reactants is 236.05 atomic mass units, and the mass of the products is 235.86 atomic mass
units. Which of the following explanations best describes the change in mass that occurs in this nuclear fission reaction?
A. Mass and energy are equivalent, and energy has been converted into mass in this reaction.
B. Mass and energy are equivalent, and mass has been converted into energy in this reaction.
C. Mass and energy are equivalent, and the missing mass is due to inaccurate laboratory measuring equipment.
D. Neutrinos that are given off in the fission reaction are undetectable, which accounts for the differences in mass of the detectable components
of the reaction.
-----------------------------------------------------------------Rutherford Gold Foil Experiment and the Planetary Model of the Atom
170. In certain scattering experiments, alpha particles bounce backward from a thin metal target. This observation led to the hypothesis that
A. alpha particles carry electric charges
B. charge is uniformly distributed throughout the atom
C. alpha particles’ kinetic energy cannot be converted to potential energy
D. the centre of the atom is very small, charged, and contains most of the atom’s mass
Use the following information to answer the next question.
171. This exercise would help students appreciate the difficulties encountered by
A. Compton in his work on wave–particle theory
B. Einstein in his work on the photoelectric effect
C. Rutherford in his work on the nucleus of the atom
D. Thomson in his work on cathode rays
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Bohr Model of the Atom (Qualitative)
172. When white light passes through a cool gas and then into a spectroscope, the spectrum produced is
A. a continuous spectrum
B. an absorption spectrum
C. a bright-line spectrum
D. an emission spectrum
Bohr Model of the Atom Problems
Use the following information to answer the next question.
173. What frequency of electromagnetic radiation is required to excite mercury atoms from energy level W to energy level Z?
A. 2.1 x 1015 Hz
B. 2.5 x 1015 Hz
C. 2.9 x 1015 Hz
D. 3.1 x 1015 Hz
-----------------------------------------------------------------Use the following information to answer the next question.
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174. When the glass cools, the ions lose both the thermal energy and the energy that was absorbed from the laser photons. The electron energy
level transition that occurs is from energy level
A. n = 3 to n = 2
B. n = 3 to n = 1
C. n = 2 to n = 1
D. n = 2 to n = 3
-----------------------------------------------------------------Unit 4 Review Questions
Use the following information to answer the next question.
175. The types of radiation taking paths X, Y, and Z are, respectively,
A. beta, alpha, and gamma
B. beta, gamma, and alpha
C. gamma, alpha, and beta
D. alpha, gamma, and beta
------------------------------------------------------------------
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Use the following information to answer the next three questions.
176. As xenon ions in the exhaust stream behind the DS1 capsule interact with other charged particles in space, the xenon ions become neutral
atoms, and in the process, emit photons. The maximum frequency of these photons, expressed in scientific notation, is b × 10w Hz. The value of b
is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
Use the following additional information to answer the next two questions.
177. If the IPS uses 81.5 kg of xenon-133 as a propellant and the launch is delayed by 26.2 days, the amount of xenon 133 that would remain is
__________ kg.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
178. The decay equation for xenon-133 is
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Use the following information to answer the next two questions.
179. Given the information above, the element that emits the lowest energy photon of visible light is
A. strontium
B. barium
C. copper
D. sodium
180. The colours are emitted by electrons that are
A. undergoing transitions to higher energy levels
B. undergoing transitions to lower energy levels
C. oscillating between energy levels
D. emitted by the nucleus
------------------------------------------------------------------
181. An experiment starts with 1.45 kg of iodine-131. After 32.2 days, 90.6 g are left. The half-life of iodine-131 is
A. 32.2 days
B. 16.1 days
C. 8.05 days
D. 4.04 days
Use the following information to answer the next four questions.
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182. The missing product, X, in the fusion reaction given above is
5
A. 2 He
4
B. 2 He
4
C. 1 H
3
D. 2 He
183. The main reason that the nuclei need to have such large kinetic energies is that
A. fusion releases large amounts of energy
B. fission must occur before fusion can occur
C. this kinetic energy is converted into nuclear energy
D. the nuclei must overcome a strong electrostatic repulsion
184. When the average kinetic energy of the nuclei in a plasma is 0.25 MeV, then the temperature is
A. 1.9 × 109 K
B. 2.9 × 109 K
C. 4.3 × 109 K
D. 1.2 × 1028 K
185. The energy of a single photon of the ultraviolet laser is
A. 7.1 × 10–19 J
B. 1.0 × 10–27 J
C. 7.1 × 10–28 J
D. 1.9 × 10–40 J
-----------------------------------------------------------------186. The absorption spectrum of hydrogen is produced when electrons
A. emit radio frequency photons
B. emit short wavelength photons
C. jump from a higher orbital to a lower orbital
D. jump from a lower orbital to a higher orbital
187. An accelerated electron with 8.77 eV of energy strikes a mercury atom and leaves the collision with 2.10 eV of energy. The maximum
frequency of light that can be emitted by the mercury atom is
A. 1.01 × 1014 Hz
B. 5.07 × 1014 Hz
C. 1.61 × 1015 Hz
D. 2.12 × 1015 Hz
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Use the following information to answer the next two questions.
188. The speed of the undeflected ionized lithium ions, Li + , as they leave the velocity selection chamber is
A. 4.25 × 104 m/s
B. 3.84 × 105 m/s
C. 8.63 × 106 m/s
D. 7.22 × 107 m/s
189. The mass of a lithium ion in beam 1, expressed in scientific notation, is b x 10–w kg. The value of b is __________.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
-----------------------------------------------------------------Use the following information to answer the next question.
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190. This distortion occurs because of the magnetic force acting on the
A. visible wavelengths of EMR
B. television circuits
C. moving electrons
D. gamma radiation
------------------------------------------------------------------
191. Which of the following types of radiation can be deflected by both electric fields and magnetic fields?
A. X-rays
B. Cathode rays
C. Photon beams
D. Electromagnetic waves
Use the following information to answer the next six questions.
192. The half-life of iodine-131 is
A. 8.0 days
B. 12.0 days
C. 16.0 days
D. 24.0 days
193. After 48.0 days the amount of iodine-131 that remains in the sample is __________ mg.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
194. The energy emitted as gamma radiation during the transmutation of an iodine-131 nucleus is
A. 3.55 × 10–45 J
B. 2.68 × 10–27 J
C. 1.24 × 10–22 J
D. 3.71 × 10–14 J
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Use the following additional information to answer the next question.
195. For the decay of iodine-131, the relationship between the magnitude of the momentum of the gamma ray photon (   ) and the magnitude
of the momentum of the beta particle (
  ) can be represented by the equation
196. The equation for this radioactive decay is
197. To protect lab technicians from harmful radiation, the equipment used in this experiment should be shielded with
A. lead to stop the radiation
B. paper to stop the particles
C. an electric field to stop the radiation
D. a magnetic field to stop the particles
-----------------------------------------------------------------198. The voltage required to stop an alpha particle with an initial speed of 5.34 × 10 4 m/s is __________ V.
(Record your three-digit answer in the numerical-response section on the answer sheet.)
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Atomic Physics MC Answers
146
147
148
149
150
151
152
153
154
155
156
157
158
159
C
D
256
D
C
D
B
D
D
B
A
D
B
A
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160
161
162
162
164
165
166
167
168
169
C
2.55
B
A
C
2937
A
3.18
1.79
B
170
171
172
173
174
175
176
177
178
179
D
C
B
A
B
B
2.92
2.55
C
A
180
181
182
183
184
185
186
187
188
189
B
C
B
D
A
A
D
C
B
1.01
190
191
192
193
194
195
196
197
198
C
A
A
1.19
D
D
D
A
29.6
159