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Physics 249 Homework 5
Due Oct 19th
1) Consider two potential steps. One from V(x<0)=0 to V(x>0)=V0 and the other from
V(x<0)=0 to V(x>0)=-V0. Consider a particle incident on this step potential from
–x with total energy E=2V0. Note this is the total energy including the potential
energy. For each calculate.
a) The wave numbers in regions two (x>0) in terms of the wave number in region 1 (x<0)
and V0.
b) The reflection coefficients.
c) The transmission coefficients.
Note since the potential energy is zero in region one the total energy is the same as the
kinetic energy.
a) The wave number for both ases i (V0) and ii (-V0) in region one is.:
𝑘1 =
√2𝑚𝐸 √4𝑚V0
=
ℏ
ℏ
in region two the wave number are
𝑘2𝑖 =
√2𝑚(2V0 − V0 ) √2𝑚V0
=
ℏ
ℏ
√2𝑚(2V0 − (−V0 ))
𝑘2𝑖 =
ℏ
=
√6𝑚V0
ℏ
b) The reflection coefficients.
2
𝑘1 − 𝑘2 2
2 − √2
𝑅𝑖 = (
) =(
) = 0.0294
𝑘1 + 𝑘2
2 + √2
2
𝑘1 − 𝑘2 2
2 − √6
𝑅𝑖𝑖 = (
) =(
) = 0.0102
𝑘1 + 𝑘2
2 + √6
c) The transmission coefficients.
𝑇𝑖 =
4𝑘1 𝑘2
8√2
=
2 = 0.9706
2
(𝑘1 + 𝑘2 )
(2 + √2)
𝑇𝑖𝑖 =
4𝑘1 𝑘2
8√6
=
2 = 0.9898
2
(𝑘1 + 𝑘2 )
(2 + √6)
Note that T+R=1 in both cases!
2) Consider potential barrier 0.5nm wide and 10eV high. Electrons accelerated through
5V approach the barrier. What percentage of electrons will tunnel through the
barrier?
−1
𝑇 = 𝑅𝑒 (
|𝐹|2
|𝐴|
2
) = [1 +
2
𝑠𝑖𝑛ℎ 𝛼𝑎
]
𝐸
𝐸
4 𝑉 (1 − 𝑉 )
0
0
Note that there was an error in the formula in the book and in my notes. The sinh should
have been squared. The formula has been corrected in my notes from lecture 16.
Solutions with either sinh2 or sinh were accepted.
−1
√2𝑚(𝑉0 − 𝐸)
𝑠𝑖𝑛ℎ (
)𝑎
ℏ
𝑇 = 1+
𝐸
𝐸
4 𝑉 (1 − 𝑉 )
0
0
[
]
2
√2𝑚𝑐 2 (𝑉0 − 𝐸)
𝑠𝑖𝑛ℎ (
)𝑎
ℏ𝑐
= 1+
𝐸
𝐸
4 𝑉 (1 − 𝑉 )
0
0
[
]
−1
2
−1
√2 ∗ 0.511𝑥106 (10 − 5)
𝑠𝑖𝑛ℎ (
) 0.5
197.3𝑒𝑉𝑛𝑚
2
𝑇 = 1+
[
= 4.23𝑥10−5 = 4.23𝑥10−3 %
4 ∗ 0.5(1 − 0.5)
]
3) Consider a infinite 3D box potential with L2=2L1 and L3=4L1. What are the quantum
numbers of the lowest degenerate energy levels? List, ordered by energy, the
quantum numbers and energies of all the levels up two the lowest energy set of
degenerate energy levels.
𝐸=
ℏ2 2
ℏ2 𝜋 2 𝑛12 𝑛22 𝑛32
(𝑘1 + 𝑘22 + 𝑘32 ) =
( + + )
2𝑚
2𝑚𝐿2 1
4 16
1, 1, 4 and 1, 2, 2
Systematically you increment up the third eigenstate comparing it to incrementing up the
second (or first) eigenstate at each step. 1, 2, 1 is smaller then 1, 1, 4 and 1, 2, 2 is
the same as 1, 1, 4. Both give fractions of ¼ and 1.
Define
ℏ2 𝜋 2
𝐸0 =
2𝑚𝐿2
for the constant term in front of the terms
1, 1, 1: 𝐸 = 1.313𝐸0
1, 1, 2 𝐸 = 1.5𝐸0
1, 1, 3 𝐸 = 1.813𝐸0
1, 2, 1 𝐸 = 2.063𝐸0
1, 1, 4 𝐸 = 2.25𝐸0
1, 2, 2 𝐸 = 2.25𝐸0
4) The radial probability distribution function for the hydrogen atom ground state can be
writer as 𝑃(𝑟) = 𝐶𝑟 2 𝑒 −2𝑟/𝑎0 , where C is a normalization constant.
a) Why is the factor of 𝑟 2 included?
b) Show that P(r) has it’s maximum value at r=a0.
c) If you considered equal infinitesimal volumes, what at what radius is the maximum
probability to find an electron?
d) Calculate the expectation value of r.
a) The probability at a give radius is integrated over a spherical shell of area 4𝜋𝑟 2.
b)
𝑑𝑃(𝑟)
2 −2𝑟
−2𝑟/𝑎0
2
= 2𝐶𝑟𝑒
− 𝐶𝑟
𝑒 𝑎0 = 0
𝑑𝑟
𝑎0
1
𝑟 − 𝑎 𝑟2 = 0
0
𝑟 = 𝑎0
−2𝑟
𝑑 2 𝑃(𝑟)
2 −2𝑟
2 −2𝑟
2 2 −2𝑟
2
𝑎
𝑎
𝑎
= 2𝐶𝑒 0 − 𝐶2𝑟 𝑒 0 − 𝐶2𝑟 𝑒 0 + 𝐶𝑟 ( ) 𝑒 𝑎0
𝑑𝑟 2
𝑎0
𝑎0
𝑎0
2
𝑑 𝑃(𝑎0 )
= 2𝐶𝑒 −2 − 8𝐶𝑒 −2 + 4𝐶𝑒 −2 = −2𝐶𝑒 −2
𝑑𝑟 2
indicating a local maxima
c)
|𝜓(𝑟)|2 = 𝐶𝑒 −2𝑟/𝑎0
Which has a maxima at r=0.
d)
4 ∞ 3 −2𝑟/𝑎
0
∫ 𝑟 𝑒
𝑎03 0
using
4 ∞ 3 −2𝑟/𝑎
4 3!
4 6 4 3
0 =
𝑎 = 𝑎0
4 = 3
3∫ 𝑟 𝑒
3
𝑎0 0
𝑎0 2
𝑎0 16 0 2
(𝑎 )
0
As expected, larger than the maximum.