Physics 220
Homework #2
Spring 2015
Due Wednesday 4/15/15
1. Photons with a wavelength λ = 410nm are used to eject electrons from a metallic
cathode (the emitter) by the photoelectric effect. The electrons are prevented from
striking the anode (the collector) by applying a stopping potential of 0.88V . What is
the work function of the metal and what is the likely composition of the metal
surface?
From the photoelectric effect, we have
hc
hc
eVstop =
−φ →φ =
− eVstop
λ
λ
⎡⎛ 6.6 × 10 −34 Js × 3 × 10 8 ms ⎞
⎤
1eV
φ = ⎢⎜
×
⎟⎠ 1.6 × 10 −19 J ⎥ − 0.88eV = 2.14eV
410 × 10 −9 m
⎣⎝
⎦
Looking up the work function at http://hyperphysics.phyastr.gsu.edu/hbase/tables/photoelec.html, the metal could be cesium at φ = 2.1eV .
2. Photons with energy E are incident on a photocathode (metal surface) and produce
electrons with a maximum kinetic energy of 2eV . When the energy of the photons is
doubled, the maximum kinetic energy of the electrons increases to 8eV . What is the
work function of the metal surface?
From the first piece of information, we can calculate the energy of the incident
photon. We have 2eV = E − φ → E = 2eV + φ . From the second piece of information
we have 8eV = 2E − φ = 2 ( 2eV + φ ) − φ = 4eV + φ → φ = 4eV .
3. When radiation of wavelength λ = 410nm is incident on a cathode with an unknown
work function, and electrons are ejected by the photoelectric effect. The minimum
voltage necessary to prevent the electrons from reaching the anode, the “stopping
potential” is unknown. Experimenters change the wavelength of the radiation and
measure the change in the stopping potential. What wavelength of radiation would
correspond to an increase in the stopping potential of 1V ?
Let λ and λ ' be the original and new wavelengths respectively. We have originally,
hc
hc
eVstop =
− φ and now eVstop + 1eV =
− φ . Solving this for λ ' we get
λ
λ'
hc
6.6 × 10 −34 Js × 3 × 10 8 ms
λ'=
=
= 308nm .
−19
−34
8 m
hc ⎛
⎞
1.6
×
10
J
6.6
×
10
Js
×
3
×
10
s
1eV +
1eV ×
⎟⎠ +
λ ⎜⎝
1eV
410 × 10 −9 m
4. Radiation with a wavelength λ = 300nm of is directed at a metallic surface. Estimate
the radiation flux (power per unit area) necessary to have an average of 1 photon per
second strike each atom on the surface of the metal.
Assuming the atom has a rough diameter of 0.3nm , we can calculate the area over
which the radiation is spread. We have A = π r 2 = π ( 0.15 × 10 −9 m ) = 7.07 × 10 −20 m 2 .
2
Next we want 1 photon per second to strike an atom, so this corresponds to an energy
(per second) of
hc
hc 6.6 × 10 −34 Js × 3 × 10 8 ms
1 photon λ
300 × 10 −9 m
Etotal = λ ×
=
=
= 6.6 × 10 −19 Js . Thus the
photon
s
s
s
intensity is S =
Energy
power
6.6 × 10 −19 Js
=
=
= 9.3 mW2 .
time × area
area
7.07 × 10 −20 m 2
5. Make a plot of eVstop versus frequency for cesium, magnesium, and platinum on the
same axis. Provide for positive values of Vstop for 0 to 5 Volts and indicate on your
graphs the visible portion of the spectrum.
Looking up the work functions at http://hyperphysics.phyastr.gsu.edu/hbase/tables/photoelec.html, we find φCs = 2.1eV , φ Mg = 3.7eV , and
φPt = 6.4eV . The plot is generated using eVstop = hf − φ for frequencies from the
infrared ( λ IR = 10 µ m ) to the ultraviolet ( λuv = 200nm ). The plot is below, where the
blue line is for cesium, the red is for magnesium, and the yellow is for platinum.
eV
6
4
2
1.0¥101 5
-2
-4
-6
1.5¥101 5
fHs- 1 L
2.0¥101 5
6. Photons with kinetic energy equal to the mass energy of an electron collide with an
π
electron at rest and scatters at an angle of . Calculate the energy of the electron
2
after the collision. Express your answer in terms of mc 2 .
Starting from the Compton wavelength formula, divide both sides by hc . We have
h
λ
+
(1− cos φ ) 1 1 (1− cos φ )
λ'
mc
. Next we evaluate the expression at
=
→
= +
hc
hc
E' E
mc 2
mc 2
0
. Now the kinetic
φ = 90 and determine the scattered photon energy to be E ' =
2
energy of the recoiling electron is the difference between the incident photon energy
mc 2 mc 2
=
and the scattered photon energy and we have KE = E − E ' = mc 2 −
. The
2
2
total energy is the sum of the electron’s rest energy and its kinetic energy. Thus the
3
energy of the recoiling electron is Ee = KE + mc 2 = mc 2 .
2
7. A photon with energy E collides with an electron at rest. Calculate the maximum
amount of kinetic energy transferred to the electron. Make a graph of KE versus E,
where the units for the energies are in electron volts.
The kinetic energy of the electron is given by KE = E − E ' , where
1 1 (1− cos φ )
Emc 2
. Combining these two equations we
= +
→ E' =
E' E
mc 2
mc 2 + E (1− cos φ )
Emc 2
have KE = E − E ' = E − 2
. The maximum energy transferred to the
mc + E (1− cos φ )
electron is when the scattered photons energy is the least and this occurs for complete
backscattering, or at an angle of φ = 180 0 . The maximum kinetic energy of the
Emc 2
2E 2
=
electron is therefore KE = E − 2
. The plot of the kinetic energy
mc + 2E mc 2 + 2E
versus the incident photon energy is shown below.
KEHMeVL
1.5
1.0
0.5
0.5
1.0
1.5
2.0
EHMeVL
8. From the three equations in class we have from conservation of total energy (1) and
conservation of momentum (2 & 3)
hc
hc
hc
1: + mc 2 =
+ Ek + mc 2 =
+E
λ
λ'
λ'
h h
2 : = cos φ + p cosθ
λ λ
h
3 : 0 = sin φ − psin θ
λ
To derive the Compton shift in wavelength, square (1) and simplify terms to get
2
⎛ hc hc ⎞
⎛ hc hc ⎞
2
4 : E = ⎜ − ⎟ + 2 ⎜ − ⎟ mc 2 + m 2 c 4 .
⎝ λ λ'⎠
⎝ λ λ'⎠
Next, square (2) and (3) and add the results. Multiply the result by c2. This produces
(5)
h2c2 h2c2
h2c2
2 2
5: p c = 2 + 2 − 2
cos φ .
λ
λ'
λλ '
Solve (4) for the expression p2c2 and set the two results equal to one another and you
obtain (6).
2
h2c2 h2c2
h2c2
⎛ hc hc ⎞
⎛ hc hc ⎞
6 : E 2 − m 2 c 4 = p 2 c 2 → ⎜ − ⎟ + 2 ⎜ − ⎟ mc 2 = 2 + 2 − 2
cos φ .
⎝ λ λ'⎠
⎝ λ λ'⎠
λ
λ'
λλ '
Solve (6) for the scattered wavelength
h
λ' = λ +
(1 − cosφ ) .
mc
(
)
9. Show that the scattering angle of the electron scattered in a Compton effect
λ sin φ
experiment is given by tan θ =
. In 1950 two scientists Cross and
h ⎞
⎛
⎜⎝ λ +
⎟ (1− cos φ )
mc ⎠
Ramey were able to detect the electrons from 2.6MeV gammas scattered to φ = 30 0 .
At what angle were they able to detect the scattered electrons?
Using conservation of momentum (2) and (3), we have dividing (3) by (2):
psin θ
hsin φ
hsin φ
tan θ =
=
=
p cosθ
⎛h h
⎞
⎛
⎞
λ ' ⎜ − cos φ ⎟
⎝ λ λ'
⎠ ⎛
⎜
⎟
h
h
h
cos φ ⎟
(1− cos φ )⎞⎟⎠ ⎜ −
⎜⎝ λ +
h
mc
⎜ λ ⎛⎜ λ +
⎟
(1− cos φ )⎞⎟⎠
⎝
⎝
⎠
mc
tan θ =
sin φ
λ sin φ
=
h ⎞
h ⎞
⎛
(1− cos φ ) ⎛⎜⎝ 1+
⎟⎠ (1− cos φ ) ⎜⎝ λ +
⎟
λ mc
mc ⎠
Using the result, we find the angle of the detected electrons, where the wavelength of
hc
the incident photons is determined from E = . Thus,
λ
λ sin φ
tan θ =
h ⎞
⎛
⎜⎝ λ +
⎟ (1− cos φ )
mc ⎠
tan θ =
4.7596 × 10 −13 m × sin 30
= 0.6145
⎛
⎞
6.6 × 10 −34 Js
−13
⎜⎝ 4.7596 × 10 m + 9.11× 10 −31 kg × 3 × 10 8 m ⎟⎠ (1− cos 30 )
s
θ = tan −1 ( 0.6415 ) = 31.6 0
Cross and Ramey actually detected them at θ = 31.30 .
10. A 1.0MeV photon collides with an electron at rest and scatters at an angle of φ = 45 0
. What are the energy of the scattered photon and the kinetic energy of the scattered
electron?
Starting as in problem #6, we’ll calculate the energy of the scattered photon and then
the kinetic energy of the scattered proton. The energy of the scattered photon is
1 1 (1− cos φ )
1
(1− cos 45 ) → E ' = 0.6357MeV . Therefore the
= +
=
+
2
E' E
mc
1MeV 0.511 MeV
× c2
c2
energy of the scattered proton is KE = E − E ' = 1MeV − 0.6357MeV = 0.3643MeV .