Chem 442 Problem Set #1 DUE: Wednesday, Nov. 5, 2008
1) McQuarrie and Simon, Problem 1-8
2) a) McQ&S, 1-19 b) The ionization potential for a Na atom is 5.15 eV. Should this be (or is this) the same as the work function? Why or why not? 3) 4) 5) 6) 7) 8) 9
Chem 442 Homework Set #7 DUE: Thursday, November 13, 2008
1) McQ&S 7-20 2 d2 k 2 b4 0 1 a) H ( ) = x + x , H ( ) = x3 + 2 2m dx 2 6 24 ( 0) = Harmonic oscillator wavefunctions.
1 E( 0) = n + 2 2 d2 ( 0) = , H (1) = b b) H 2 2m dx
a < x < a 2
( 0) =
E( 0
Chem 442 Homework Set #8 DUE: Friday, November 14, 2008
1) 2) 3) McQ&S, 8-22 McQ&S, 8-32 McQ&S, 8-47. Instead of giving a suggestion for the trend, use the following Hamiltonian to determine, AS.O., the spin-orbit coupling constant for F, Cl, Br, and I. (
Chem 442 Homework Set #8 DUE: Friday, November 14, 2008
1) McQ&S, 8-22 In this problem we are only concerned with the spin part of the wavefunciton: 1 (1) (1) 1 = ( (1) ( 2 ) (1) ( 2 ) ) 2 ( 2) ( 2) 2
=
S z ,total = S z1 + S z 2 = S z1 + S z 2
(
)
(
) 12
Chem 442 Homework Set #9 DUE: Tuesday, November 18, 2008
1) McQ&S, 9-15 C2
* u 2 p * g 2p
C2
* u 2 p * g 2p
g2p u 2 p
g2p u 2 p
* u 2s
* u 2s
g 2s
g 2s
Bond Order= C2 : B.O. = 2
1 ( number of bonding electrons - number of antibonding electrons ) 2
C2 :
Chem 442 Homework Set #10 DUE: Wednesday, November 19, 2008
1) 2) 3) 4) McQ&S, 10-18 McQ&S, 10-20 McQ&S, 10-23 The lowest energy electronic transitions in ethene (C2H4) and benzene (C6H6) occur at 28,700 and 29,510 cm-1, respectively. Using Hckel theory,
Chem 442 Homework Set #10 DUE: Wednesday, November 19, 2008
1) McQ&S, 10-18
CO2
Linear
CO2 +
Linear
CF2
Bent
2)
3)
McQ&S, 10-20 Since this orbital has no dependence on the bond angle, it must be a core orbital. So, the 1a1 is just the 1s orbital on X. Sin
Chem 442 Homework Set #12 Friday, November 21, 2008 1) 2) McQ&S, 15-9 Assume that the infrared dipole transition moment for HCl is the same as the value in problem 15-9. Calculate the radiative lifetime for the v =1 v = 0 transition at 2980cm-1. This is o
Chem 442 Homework Set #7 DUE: Thursday, November 13, 2008
1) 2) 3) 4)a) McQ&S 7-20 McQ&S 7-22 McQ&S 7-23 Using = c1 f1 + c2 f 2 where f1 = x 2 ( L x ) and f 2 = x ( L x ) as a trial function, apply the variational treatment to calculate an upper bound to
Chem 442 Homework Set #6 DUE: Wednesday November 12, 2008
1) McQ&S 6-46
The magnitude of the spiltting is B Bz m.
B = 9.2740154 1024 J T 1 (Front cover of book - its denoted as B )
So, the magnitude of the splitting is B Bz m = ( 9.2740154 10 24 J T 1 )
Chem 442 Problem Set #1 - Solutions DUE: Wednesday, Nov. 5, 2008
1) (McQ&S 1-8)
Using the Wien displacement law (eqn 1.4) or Planck's law (1.5):
maxT = 2.90 103 m K
2.90 103 m K max = = 2.90 1010 m 7 10 K Considering sig figs, we have max = 3 1010 m = 0.3
Chem 442 Homework Set #2 DUE: Thursday, November 6, 2008 1) 2) 3) a) b) McQ&S, 3-3 McQ&S, 3-6 McQ&S, 3-28. In addition, please explain the following points: Why is n = 0 a valid (i.e. non-trivial) solution? For n 0, En = E-n (i.e. n() and -n() have the sa
Chem 442 Homework Set #2 - Solutions DUE: Thursday, Nov. 6. 2008
1)
The eigenvalue equation is Af ( x ) = cf ( x ) where c is a constant.
(McQ&S 3-3)
2 ( x ) = d cos x = 2 cos x = 2 f ( x ) ; eigenvalue = 2 a) Af dx 2 d b) Af ( x ) = eit = i eit = i f (
Chem 442 Homework Set #3 DUE: Friday, November 7, 2008
1) a) McQ&S, 3-26 b) Assume there is an electron in this box where a=5.00. Calculate the frequency of a photon emitted if the electron makes a transition from the first excited state to the ground sta
Chem 442 Homework Set #3 - Solutions DUE: Friday, November 7, 2008
1) a) McQ&S, 3-26
The energy levels for the particle in the 3-D box are: Enx ny nz Enx ny nz Enx ny nz
2 nz 2 h 2 nx 2 n y = + + 8m a 2 b 2 c 2
plug in a=b=1.5c
2 nz 2 h 2 nx 2 n y = + +
Chem 442 Homework Set #4 DUE: Saturday, November 8, 2008
1) In lecture 9, I introduced two new states for the particle on a ring:
n+ =
a)
1
cos ( n ) and n =
1
sin ( n )
b) c) d) 2) a) b) c) 3) 4) 5)
Prove statements A and B on slide 5 of lecture nine. T
Chem 442B Homework Set #4 - Solutions DUE: Saturday, Nov. 8, 2008
1)a)
2 d2 1 d n cos ( n ) = sin ( n ) 2 I d 2 2 I d 22 n1 cos ( n ) = En n + = 2I 2 2
H n+ =
H n =
2 d2 1 dn sin ( n ) = cos ( n ) 2 I d 2 2 I d 22 n1 sin ( n ) = En n = 2I
b)
2
0
n +
Chem 442B Homework Set #5 DUE: Tuesday, November 11, 2008
1) Determine the total probability that a harmonic oscillator in the first excited state is in the classically forbidden region. Hints: In this case xm 1/ . First figure out xm in terms of . Set up
Chem 442B Homework Set #5 - Solutions DUE: Tuesday, November 11, 2008
1)
4 6 2 x2 / 2 1 ( x ) = (pg 170 McQ&S) xe xm is the value of x where the potential energy = the total energy 1 E1 = 1 + 2 1 1 V ( x ) = kx 2 = m 2 x 2 2 2 3 1 = m 2 xm 2 2 2 m 3 3 xm
Chem 442 Homework Set #6 DUE: Wednesday, November 12, 2008
1) 2) 3) McQ&S 6-46 McQ&S 6-47 In this problem and the next, we will find the approximate solutions for the quartic oscillator using the variational method. The Hamiltonian for the quartic 2 d2 =
Chem 442 Homework Set #12 Friday, November 21, 2008 1)
A= 1
McQ&S, 15-9
R
c = 1 = 6.25 108 s 1 9 1.6 10 s
g1 = 1, g 2 = 3 2.998 108 m / s v= = = 2.461 1015 s 1 9 121.8 10 m 3 hc3 g 2 = 03 3 2A 16 v g1 = 3 ( 8.854 1012 C 2 J 1 m 1 )( 6.626 1034 J s )( 2.99