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Physics 322: Modern Physics
Spring 2015
Problem Set #8 (Having Tunneled Through Spring Break…)
Due Friday, March 27 in lecture.
ASSUMED READING: Before starting this homework, you should read all of
Chapter 6 of Harris’ Modern Physics (focus on Sections 1 and 2)
1. [Harris 6.1 (tweaked)] Could the
situation depicted in the diagram to the right
represent a particle in a bound state? Clearly
explain your reasoning. HINT: How would
your answer change if the potential barriers
were 2x as wide? 200x as wide? 2. [Harris 6.20 (modified)] Particles of energy ! E are incident from the
left, where U(x)=0, and at the origin encounter an abrupt drop in potential
energy, whose depth is ! −3E .
a. Sketch this potential including the total energy of the particles on
the same scale.
b. Classically, what would the particles do. Specifically, what would
happen to their kinetic energy as they move and why?
c. Apply quantum mechanics. Assuming an incident wave of the form
! ψ inc ( x ) = 1eikx (where the normalization constant has been given the
simple value of 1), determine completely the wave function
everywhere that describes this particle stream, including numeric
values for the multiplicative constants.
d. What is the probability that incident particles will be reflected? 3. [Harris 6.15 (tweaked)] Calculate the reflection probability for a 5eV
electron encountering a step in which the potential drops by 2eV. Clearly
explain your reasoning. HINT: If you are doing a lot of math, you are
approaching the problem incorrectly. The mathematical heavy lifting was
done in Harris, you just need to explain its application here. – Page 1! of 3! –
Physics 322: Modern Physics
Spring 2015
4. [Harris 6.9] In the wide-barrier transmission probability of equation
(6-18), E ⎛
E ⎞ −2 L 2 m (U0 − E ) /!
! T ≅ 16 ⎜ 1 − ⎟ e
U0 ⎝ U0 ⎠
the coefficient multiplying the exponential, ! 16
E ⎛
E⎞
1
−
, is often
U 0 ⎜⎝ U 0 ⎟⎠
omitted. When is this justified, and why? NOTE: Understanding the
answer to this may simplify the next problem. 5. [Harris 6.32] Jump to Jupiter. The gravitational potential energy of a 1
kg object is plotted versus position from Earth’s surface to the surface of
Jupiter. Mostly it is due to the Sun, but there are downturns at each end,
due to the attractions of the two planets. Make a crude approximation
that this is a rectangular barrier
o f w i d t h 6 x 1 011 m a n d
approximate height (in terms of
energy per kilogram lifted) of
4×108 J/kg. Your mass is 65
kg, and you launch yourself
from Earth at an impressive 4
m/s. What is the probability
that you can jump to Jupiter? MATH NOTE: Since most of your calculators can’t work with numbers
this small, to convert your exponential into a power of 10, use the fact that
log e N
! log10 N =
. log e 10
6. [Harris 6.35] Fusion in the Sun: Without tunneling, our Sun would
fail us. The source of its energy is nuclear fusion, and a crucial step is
the fusion of a light-hydrogen nucleus, which is a proton, and a heavyhydrogen nucleus, which is of the same charge but twice the mass. When
these nuclei get close enough, their short range attraction via the strong
force overcomes their Coulomb repulsion. This allows them to stick
together, resulting in a reduced total mass/internal energy and a
consequent release of kinetic energy. However, the Sun’s temperature is
simply too low to ensure that nuclei move fast enough to overcome their
repulsion.
a. By equating the average thermal kinetic energy that the nuclei
3
would have when distant, ! kBT , an the Coulomb potential energy
2
– Page 2! of 3! –
Physics 322: Modern Physics
Spring 2015
they would have when 2 fm apart, roughly the separation at which
they stick, show that a temperature of about 109 K would be needed.
For full credit, explain why this is a good “classical” guess on the
temperature necessary for fusion.
b. The Sun’s core is only about 107 K. If
nuclei can’t make it “over the top,” they
must tunnel. Consider the following
model, illustrated in the figure: One
nucleus is fixed at the origin, while the
other approaches from far away with
energy E. As r decreases, the Coulomb
potential energy increases until the
separation r is roughly the nuclear
radius rnuc, whereupon the potential
energy is Umax and then quickly drops
down into a very deep “hole” as the
strong-force attraction takes over. Given
that E≪Umax, the point b, where the tunneling must begin, will be
very large compared with rnuc, so we approximate the barrier’s
width L as simply b. Its height. U0, we approximate by the Coulomb
potential evaluated at b/2. Finally, for the energy E, which fixes b,
3
let us use ! 4 × kBT , which is a reasonable limit, given the natural
2
range of speeds in a thermodynamic system. Combining these
approximations, show that the exponential factor in the widebarrier tunneling probability (equation 6-18) is ⎡ −e2
4m ⎤
! exp ⎢
⎥.
⎣ ( 4π ε0 ) ! 3kBT ⎦
c. Using the proton mass for m, evaluate this factor for a temperature
of 107 K (on the order of the temperature of the Sun’s core). Then
evaluate it at 3000K, the temperature of an incandescent filament
or hot flame, and rather high by Earth standards. Discuss the
consequences. – Page 3! of 3! –