Download Chapter 6 Quantum Mechanics in One Dimension. Home

Chapter 6
Quantum Mechanics in One
Dimension. Home Work Solutions
6.1
Problem 6.9 (In the text book)
The nuclear potential that binds protons and neutrons in the nucleus of an atom is often
approximated by a square well. Imagine a proton confined in an infinite square well of length
10−5 nm, a typical nuclear diameter. Calculate the wavelength and energy associated with
the photon that is emitted when the proton undergoes a transition from the first excited
state (n = 2) to the ground state (n = 1). In what region of the electromagnetic spectrum
does this wavelength belong?
Solution
The energy levels of a particle in an infinite square well width L are given by:
En =
n2 h2
8mL2
n = 1, 2, 3, · · ·
where m is the mass the particle.
The energy given off in the transition is:
∆E = E2 − E1 =
3h2
3(hc)2
=
8mp L2
8mp c2 L2
where mp is the mass of the proton,
∆E =
3 × (1.24 × 103 )2
= 6.145 M eV
8 × 938.3 × 106 × (10−5 )2
2
CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. HOME WORK
SOLUTIONS
this energy will be taken up by a photon, with a wave length λ given by:
hc
1.24 × 10−3
=
= 2.018 × 10−4 nm
∆E
6.145
This is the gamma-ray region of the electromagnetic spectrum
λ=
Physics 205:Modern Physics I, Chapter 6
Fall 2004
Ahmed H. Hussein
6.2. PROBLEM 6.12 (IN THE TEXT BOOK)
6.2
3
Problem 6.12 (In the text book)
A ruby laser emits light of wavelength 694.3 nm. If this light is due to transitions from the
n = 2 state to the n = 1 state of an electron in a box, find the width of the box.
Solution
In a box, an electron is allowed to have the following energies:
n2 h2
8mL2
and the transition energy is related to the emitted photon wavelength by
En =
∆E =
h2 c2
hc
(22 − 12 ) =
2
2
8mc L
λ
we can then find the length of the box from;
r
r
3hcλ
3 × 1.24 × 103 × 694.3
=
= 0.795 nm
L=
8mc2
8 × 511 × 103
Physics 205:Modern Physics I, Chapter 6
Fall 2004
Ahmed H. Hussein
4
CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. HOME WORK
SOLUTIONS
6.3
Problem 6.21 (In the text book)
Sketch the wavefunction ψ(x) and the probability density |ψ(x)|2 for the n = 4 state of a
particle in a finite potential well.
Solution
Canceled
Physics 205:Modern Physics I, Chapter 6
Fall 2004
Ahmed H. Hussein
6.4. PROBLEM 6.25 (IN THE TEXT BOOK)
6.4
5
Problem 6.25 (In the text book)
Show that the oscillator energies in Equation 6.29 correspond to the classical amplitudes
r
(2n + 1)~
An =
mω
Solution
The total energy of a classical particle oscillating under the effect of a restoring force is:
1
1
E = mv 2 + kx2
2
2
where k is the force constant, m is the mass of the particle, v is particle’s velocity, and x
is the position of the particle relative to the equilibrium position. At the extreme points
x = ±A the total energy becomes totally potential, i.e.
1
E = kA2
2
with k = mω 2 , then
1
E = mω 2 A2
2
If the energy is quantized as in the quantum oscillator, then
1
1
E = n+
~ω = mω 2 A2
2
2
for a particular n the amplitude is given by:
r
An =
Physics 205:Modern Physics I, Chapter 6
(2n + 1)~
mω
Fall 2004
Ahmed H. Hussein
6
CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. HOME WORK
SOLUTIONS
6.5
Problem 6.33 (In the text book)
(a) What value do you expect for < px > for the quantum oscillator? Support your answer
with a symmetry argument rather than a calculation.
(b) Energy principles for the quantum oscillator can be used to relate < p2x > to < x2 >.
Use this relation, along with the value of < x2 > from Problem 32, to find < p2x > for
the oscillator ground state.
(c) Evaluate ∆px , using the results of (a) and (b).
Solution
(a) In quantum oscillator, the re is no preference for the direction of the motion, so the
particle will spend equal times moving to the left of the center and to the right. As a
result the average value of the momentum < px > = 0.
(b) We can find the < p2x > from the energy principles by writing < E >:
<E>=
p2x
2m
2
p
+ < U (x) > = x + < U (x) >
2m
In a quantum oscillator, the energy is sharp and in the ground state the energy is:
1
< E >= E◦ = ~ω
2
Using U (x) = 12 mω 2 x2 and take the value of < x2 > = ~/2mω (see solution of problem
6.32 in Selected Solutions of chapter 6 on the web) we get:
1
1
< U (x) > = mω 2 < x2 > = ~ω
2
4
Now, < p2x > becomes:
<
p2x
1
1
1
> = 2m(< E > − < U (x) >) = 2m(E◦ − ~ω) = 2m( ~ω− ~ω) = 2m
4
2
4
~ω
4
1
= m~ω
2
(c) ∆px is given by;
r
∆px =
p
Physics 205:Modern Physics I, Chapter 6
< p2x > − < px >2 =
Fall 2004
1
m~ω − 0 =
2
r
m~ω
2
Ahmed H. Hussein