Download Chapter 5 Physics 205 Solution of Home Work Problems

Chapter 5
Physics 205 Solution of Home Work
Problems
5.1
Problem 5.18
Find points of maximum and minimum probability density for the nth state of a particle in
a one-dimensional box. Check your results for the n = 2 state.
Solution
The wave function for a particle one-dimension box of width L is given by:
nπx ψn = A sin
L
The probability density is:
P (x) = |ψn |2
2
= A sin
2
nπx L
The maximum value of P (x) is one and occurs when:
nπx
π 3π 5π
=
, , ,···
L
22 2 1
= π m+
2
L
1
x =
m+
n
2
m = 0, 1, 2, 3, · · · n
2
CHAPTER 5. PHYSICS 205 SOLUTION OF HOME WORK PROBLEMS
Similarly the minimum value of P (x) is zero and occurs when:
nπx
= 0, π, 2π, 3π · · ·
L
= mπ
Lm
x =
n
m = 0, 1, 2, 3, · · · n
For n = 2 state, the maximum occurs when m = 0, 1, 2 at:
x=
L 3L 5L
,
,
.
4 4 4
and the minimum occurs when m = 0, 1, 2 at:
1
x = 0, L, L.
2
5.2. PROBLEM 5.23
5.2
3
Problem 5.23
Consider a square well having an infinite wall at x = 0 and a wall of height U at x = L
..
(Figure 5.23). For the case of E < U , obtain solutions to the Shrodinger equation inside the
well (0 ≤ x ≤ L) and in the region beyond (x > L) that satisfy the appropriate boundary
conditions at x = 0 and x = ∞. Enforce the proper matching conditions at x = L to find an
equation for the allowed energies of this system. Are there conditions for which no solution
is possible? Explain.
To infinity
E
U
L
0
Figure 5.23: One dimension box with infinite wall on one side and a finite one on the opposite
side.
Solution
The particle is free inside the well, so the wave function is given by:
ψ(x) = A sin kx + B cos kx
0≤x≤L
p
Where k = 2mE/~2 . The infinite wall at x = 0 requires that ψ(0) = 0 which means that
B = 0.
..
Beyond x = L, the potential energy U (x) = U and the Shrodinger equation becomes:
d2 ψ
2m
=
{U − E} ψ(x)
2
dx
~
4
CHAPTER 5. PHYSICS 205 SOLUTION OF HOME WORK PROBLEMS
for E < U the solution of the last equation is exponential:
ψ(x) = Ce−αx + De+αx
x>L
p
Where α = 2m(U − E)/~2 . To keep ψ finite at x = ∞ we must take D = 0. At x = L,
continuity of ψ and dψ/dx requires:
A sin kL = Ce−αL
kA cos kL = −αCe−αL
Dividing the above two equation will give us an equation for the allowed particle energies:
k cot kL = −α
Using k 2 + α2 = 2mU/~2 and cot2 θ + 1 = 1/ sin2 θ we get:
r
2mU
k cot kL = −
− k2
~2
2mU L2
(kL)2 cot2 kL =
− (kL)2
~2
2mU L2
(kL)2 [cot2 (kL) + 1] =
2
r~
kL
2mU L2
=
sin kL
~2
The last equation determines the allowed particle k or E. Since (θ/ sin θ) can never be
smaller that unity for any value of θ, then there will be no bound state energies when:
2mU L2
<1
~2