Solutions to Homework Problems in UPC Chapter 6
January 22, 2015
1a) To calculate the force constant we employ Eq. 6.5
=
1
=
c
2c
f
m
therefore
f = 4 2 mc2 2
= 4(3.142)2 (1.7 1027 kg)(3 1010 cm/s)2 (3
Solutions to Homework Problems in UPC Chapter 6
January 12, 2018
1a) To calculate the force constant we employ Eq. 6.5
1
= =
c
2c
r
f
m
therefore
f = 4 2 mc2 2
= 4(3.142)2 (1.7 1027 kg)(3 1010 cm/s)2
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Solutions to Homework Problems in UPC Chapter 6
January 28, 2015
2
5) a) (x) = ex , this is a unnormalized wave-function.
1
exp(-x*x)
0.9
0.8
0.7
Psi(x)
0.6
0.5
0.4
0.3
0.2
0.1
0
-2
-1.5
-1
-0.5
0
x
0
Solutions to Homework Problems in UPC Chapter 6
January 25, 2018
3)
a) We need to test whether or not a given wavefunction, , is an eigenfunction of the momentum operator,
px = i~
d
dx
I)
(x) = A sin(
Solutions to Homework Problems in UPC Chapter 7
January 31, 2018
1)
r
n (x) =
nx
2
sin
b
b
|n|2(x)
a)
0
0.2
0.4
0.6
0.8
1
x/b
Figure 1: Particle-in-a-box probability density functions. The lowest cu
CHM 374
EXAM I
3 February 2016
Name: CPA-7
This exam should have a total of 10 numbered pages,
including the back pages which contain reference material.
Show all your work to receive full credit for
Solutions to Homework Problems in UPC Chapter 7
February 4, 2015
1)
2
nx
sin
b
b
n (x) =
|n|2(x)
a) b) From Fig. 1 we can note that for n = 2 the p.d.f. has two lobes each of 1/2 probability, and sin2
Solutions to Homework Problems in UPC Chapter 8
March 12, 2015
8) Li2+ is a one-electron (hydrogen-like) ion with a nuclear charge of Z=3.
a) We could use Eqs. 8.7 and 8.8 to obtain the appropriate ex
Solutions to Homework Problems in UPC Chapter 7
February 26, 2015
14) We need to make use of the fact that [x, px ] = i to prove that [, a ] = 1.
a
Let c = (2m )1/2 , we know that:
and a = c(mx ix )
Solutions to Homework Problems in UPC Chapter 7
February 12, 2015
6) The ground state and excited state congurations are shown in the following gure. The LUMO and
n=-2
n=+2
n=-2
n=+2
n=-1
n=+1
n=-1
n=
Solutions to Homework Problems in UPC Chapter 8
March 6, 2015
1a) The energy is obtained from the formula
En =
hcRH
(6.63 1034 J s)(3.0 108 m/s)(1.0974 107 m1 )
=
n2
n2
En =
2.18 1018 J
n2
and
L2 =