Download FLORIDA INTERNATIONAL UNIVERSITY

CHM 3411 – Problem Set 2
Due date: Wednesday, January 27th
Do all of the following problems. Show your work.
"I do not like it, and I am sorry I ever had anything to do with it." - Erwin Schrödinger (about quantum mechanics)
1) Find the quantum mechanical operators for the following quantities.
a) v (velocity)
b) p2 (square of momentum)
2) We discussed the one dimensional particle in a box problem in class. The potential for this system is
V(x) =  for x  0 or x  L
V(x) = 0 for 0 < x < L
The wavefunctions that are solutions to the TISE for this system and the corresponding energy levels are
n(x) = (2/L)1/2 sin(nx/L) inside the box
En = n2h2/8mL2
n(x) = 0 outside the box
n = 1, 2, 3, …
a) Write the wavefunctions corresponding to the n = 2 and n = 3 states of the above system.
b) Find P(L/4 < x < L/2), the probability of the particle being found between L/4 and L/2, for the n = 2 and
n = 3 states.
c) Show that the n = 2 and n = 3 states are orthonormal, that is, show the following
0L 2* 2 dx = 0L 3* 3 dx = 1 ; 0L 2* 3 dx = 0
d) Is f(x) = exp(ikx), where k is a positive constant, an eigenfunction of the Hamiltonian for the particle in a
box for the region where V = 0? If so, is it an acceptable solution to the TISE for the system? Justify your answer.
e) Find a general formula for the wavelength of light that an electron would need to absorb to move from
the n = 1 to the n = 2 state of a particle in a box. Find the value for wavelength (in nm) when the box size is 0.100
nm.
3) The possible values for energy for the particle in a cube are
E(nx,ny,nz) = (nx2 + ny2 + nz2) E0 ; where E0 = h2/8mL2
nx = 1, 2, 3, …
ny = 1, 2, 3, …
(3.1)
nz = 1, 2, 3, …
Find all of the energy levels for the above system with E  40 E0. List the quantum numbers for each state (nx,ny,nz).
Plot your results in an energy level diagram. For each possible value for energy give the degeneracy of the level.
Also do the following exercises from Atkins:
9.3 a
9.7 b
Solutions
1)
a) Since p = mv, v = p/m, and so
v = (1/m) (- i d/dx) = - (i/m) d/dx
b) p2 = ( - i d/dx)2 = - 2 d2/dx2
2)
a) Based on the general formula for the wavefunctions, the n = 2 and n = 3 solutions are (inside the box)
n=2
(2/L)1/2 sin(2x/L)
n=3
(2/L)1/2 sin(3x/L)
b) The general integral we will need is
 [sin(ax)]2 dx = x/2 - (1/4a) sin(2ax)
For n = 2
P = 0L/4 [(2/L)1/2 sin(2x/L)]2 dx = (2/L) 0L/4 [(sin(2x/L)]2 dx Let a = 2/L, then
= (2/L) { x/2 - (L/8) sin(4x/L) }0L/4
= (2/L) { (L/8) - (L/8) sin() } = 1/4 = 0.2500...
For n = 3
P = 0L/4 [(2/L)1/2 sin(3x/L)]2 dx = (2/L) 0L/4 [(sin(3x/L)]2 dx Let a = 3/L, then
= (2/L) { x/2 - (L/12) sin(6x/L) }0L/4
= (2/L) { (L/8) - (L/12) sin(3/2) } = 1/4 + (1/6)  0.3031...
c) There are three integrals to consider (since the particle in a box wavefunctions are real, * = ). The
general integral needed to do the first two integrals is given in part b of the problem.
0L 2* 2 dx = 0L [(2/L)1/2 sin(2x/L)]2 dx = (2/L) 0L [(sin(2x/L)]2 dx Let a = 2/L, then
= (2/L) { x/2 - (L/8) sin(4x/L) }0L
= (2/L) { (L/8) - (L/8) sin(4) } = 1
0L 3* 3 dx = 0L [(2/L)1/2 sin(3x/L)]2 dx = (2/L) 0L [(sin(3x/L)]2 dx Let a = 2/L, then
= (2/L) { x/2 - (L/8) sin(6x/L) }0L
= (2/L) { (L/8) - (L/8) sin(6) } = 1
For the third integral
0L 2* 3 dx =
We need the following general integral
=  sin(mx) sin(nx) dx = sin(m-n)x - sin(m+n)x
m2  n2
2(m-n)
So
2(m+n)
0L 2* 3 dx = 0L [sin(2x/L) sin(3x/L)] dx
Let m = 2/L and n = 3/L, then
= { sin(-x/L) - sin(5x/L) }0L
(-2/L)
(10/L)
= 0 , (since sin(-) = sin(5) = sin(0) = 0 )
d) The Hamiltonian, inside the box, is
H = - (2/2m) d2/dx2
and so
H = - (2/2m) d2/dx2 eikx = - (2/2m) (ik)2 eikx = 2k2/2m eikx
so yes, it is an eigenfunction, with eigenvalue = 2k2/2m
Howeverm it is not an acceptable solution because it cannot satisfy the boundary conditions for the particle
in a box, namely, that (0) = (L) = 0. For example, at x = 0
(0) = eik0 = e0 = 1
Since there is no finite normalization constant we can multiply the wavefunction by to make it equal to zero at x = 0,
this function is not an acceptable solution to the particle in a box.
This is an important point. To be an acceptable solution, a function must be an eigenfunction of H and
satisfy all of the boundary conditions for the problem.
e) The general formula for the energy is En = n2 E0, where E0 = h2/8mL2.
For a transition n = 1  2, the difference in energy is
E = E2 - E1 = 4 E0 - E0 = 3 E0.
But
E = hc/ , and so  = hc/E = hc/(3E0) = 8mcL2/3h.
For L = 0.100 nm (approximately the size of a hydrogen atom)
 = 8 (9.11 x 10-31 kg) (2.998 x 108 m/s) (0.100 x 10-9 m)2 = 1.1 x 10-8 m = 11 nm
3 (6.626 x 10-34 J.s)
3) We will first list all of the states systematically, with nx  ny  nz.
nx,ny,nz
nx2+ny2+nz2
nx,ny,nz
nx2+ny2+nz2
(1,1,1)
3
(4,4,1)
33
(2,1,1)
6
(2,2,2)
12
(3,1,1)
11
(3,2,2)
17
(4,1,1)
18
(4,2,2)
24
(5,1,1)
27
(5,2,2)
33
(6,1,1)
38
(3,3,2)
22
(2,2,1)
9
(4,3,2)
29
(3,2,1)
14
(5,3,2)
38
(4,2,1)
21
(4,4,2)
36
(5,2,1)
30
(3,3,3)
27
(3,3,1)
19
(4,3,3)
34
(4,3,1)
26
(5,3,1)
35
On the next page we give a plot showing the location and numer of all the states with E  40 E0. We then make a
few additional comments about this problem.
|
|
| 38 E0. (6,1,1) (1,6,1) (1,1,6) (5,3,2) (5,2,3) (3,5,2) (3,2,5) (2,5,3) (2,3,5)
|
g=9
| 36 E0. (4,4,2) (4,2,4) (2,4,4)
| 35 E0. (5,3,1) (5,1,3) (3,5,1) (3,1,5) (1,5,3) (1,3,5)
| 34 E0. (4,3,3) (3,4,3) (3,3,4)
| 33 E0. (4,4,1) (4,1,4) (1,4,4) (5,2,2) (2,5,2) (2,2,5)
|
|
| 30 E0. (5,2,1) (5,1,2) (2,5,1) (2,1,5) (1,5,2) (1,2,5)
| 29 E0. (4,3,2) (4,2,3) (3,4,2) (3,2,4) (2,4,3) (2,3,4)
|
| 27 E0. (5,1,1) (1,5,1) (1,1,5) (3,3,3)
| 26 E0. (4,3,1) (4,1,3) (3,4,1) (3,1,4) (1,4,3) (1,3,4)
|
| 24 E0. (4,2,2) (2,4,2) (2,2,4)
|
| 22 E0. (3,3,2) (3,2,3) (2,3,3)
| 21 E0. (4,2,1) (4,1,2) (2,4,1) (2,1,4) (1,4,2) (1,2,4)
|
| 19 E0. (3,3,1) (3,1,3) (1,3,3)
| 18 E0. (4,1,1) (1,4,1) (1,1,4)
| 17 E0. (3,2,2) (2,3,2) (2,2,3)
|
|
| 14 E0. (3,2,1) (3,1,2) (2,3,1) (2,1,3) (1,3,2) (1,2,3)
|
| 12 E0. (2,2,2)
| 11 E0. (3,1,1) (1,3,1) (1,1,3)
|
| 9 E0. (2,2,1) (2,1,2) (1,2,2)
|
|
| 6 E0. (2,1,1) (1,2,1) (1,1,2)
|
|
| 3 E0. (1,1,1)
|
|
.
We can count the number of states in a particular energy range
36-40 E0.
31-35 E0.
26-30 E0.
21-25 E0.
16-20 E0.
11-15 E0.
6-10 E0.
1-5 E0.
12 states
15 states
22 states
12 states
9 states
10 states
6 states
1 state
These data are plotted below as # of states vs E1/2.
g=3
g=6
g=3
g=6
g=6
g=6
g=4
g=6
g=3
g=3
g=6
g=3
g=3
g=3
g=6
g=1
g=3
g=3
g=3
g=1
It can be shown that for sufficiently large values of E that # of states ~ E1/2.
Exercise 9.3 a
For the n = 1 state of the particle in a box
(x) = (2/L)1/2 sin(x/L)
<p> = 0L [(2/L)1/2 sin(x/L)] (-i d/dx) [(2/L)1/2 sin(x/L)] dx
= ( -i) (2/L) (/L) 0L sin(x/L) cos(x/L) dx
The general form for this integral is
 sin(ax) cos(ax) dx = (1/2a) [sin(ax)]2
Let a = /L, then
= - (2i/L2) (L2/2) [sin(x/L]0L = 0
<p2> = 0L [(2/L)1/2 sin(x/L)] (-2 d2/dx2) [(2/L)1/2 sin(x/L)] dx
= ( -2) (2/L) (/L)2 (-1) 0L [sin(x/L)]2 dx
The general form for this integral has been given in a previous problem. Let a = /L, then
= (222/L3) { x/2 - (L/4) sin(2x/L }0L
= (222/L3) (L/2) = 22/L2 = h2/4L2
Exercise 9.7b
The problem is worded somewhat ambiguously, but it is saying that
E = n2 E0, where E0 = h2/8mL2
m(N2) = (28.0 x 10-3 kg/mol) (1 mol/6.022 x 1023) = 4.65 x 10-26 kg
So
E0 =
(6.626 x 10-34 J.s)2
= 1.18 x 10-42 J
8 (4.65 x 10-26 kg) (1.00 m)2
Since E = (3/2) kT = (3/2) (1.381 x 10 -23 J/K) (300. K) = 6.21 x 10-21 J
Then n2 = E/E0 = (6.21 x 10-21)/(1.18 x 10-42) = 5.26 x 1021
n = (n2)1/2 = 7.25 x 1010
The separation between adjacent energy levels is
E = E(n+1) - E(n) = (n+1)2E0 - n2E0 = [(n2 + 2n + 1) - n2] E0 = (2n+1) E0
For the above value of n, E = [2 (7.25 x 1010) + 1] (1.18 x 10-42 J) = 1.71 x 10-31 J
Note that since E is small compared to E the energy levels are essenially continuous.
Finally, the de Broglie wavelength is
deBroglie = h/mv
But E = p2/2m, and so p = mv = (2mE)1/2 = [2 (4.65 x 10-26 kg) (6.21 x 10-21 J)]1/2 = 2.40 x 10-23 kg.m/s
deBroglie = h/mv = (6.626 x 10-34 J.s)/(2.40 x 10-23 kg.m/s) = 2.76 x 10-11 m
Particles behave classically (that is, as particles) on size scales large compared to the de Broglie wavelength. Since
the de Broglie wavelength is so small the particle will behave classically.