Chem 1410
Problem Set 5
Solution Key
(0) a) In region II (inside the box), the Schrdinger Eq. reads:
2 ( x)
= E ( x) [Region II]
2m x 2
The solutions to this diff. eq. are (modulo normalization facto
CHEMISTRY 1480 Exam 2, Spring 2015
1a. (4 points)
Electron-electron repulsion. Both electrons are moving so it is difficult to analytically solve for this
interaction. The distance between the two par
1.
There is a limit to the accuracy in which an object can be measured. Mathematically it is
4 4 . This is the error in measuring the position and velocity.
As m increases, decrease and become practi
CHEMISTRY 1480 Exam 4, Spring 2015
Multiple Choice: Circle the best answer (4 points each)
1. Which partition function contributes the most to entropy of a molecular gas?
A)
B)
C)
D)
E)
translation
ro
Sept. 12, 2007
Chem. 1410
Problem Set 1, Solutions
(1) (4 points) Note: since = c / , then, given a small change in wavelength of about a
central value of , the corresponding frequency change is = (d
Nov. 26 2007
Chem. 1410
Problem Set 10: solution to Problem 2
a) Using the given functions S ( R), J '( R), K '( R) , a plot of E ( R) 1/ 2 (which reaches the
asymptotic value of 0 as R
) vs. R is sho
Chem. 1410, Problem Set 9; solution key.
1) a) The two basis functions are orthogonal, and are unit-normalized. That is,
dx ( x) ( x) =
i
j
ij
for
i = 1, 2 . [Recall that ij is the Kronecker delta:
P9.1)
(1) Angular momentum states of a hydrogenic atom
Since the energy eigenfunction
and L , then:
nlm
(r , , ) of a hydrogenic atom is also an eigenfunction of L2
z
a) For a 2p orbital (l=1): L =
Chem 1410
PS 6, Solution
key
(0) i) For ICN N=#atoms=3, and, since it is a linear molecule nv =# vibrational degrees of
freedom (normal modes of vibration)= 3N-5 = 4.
ii) For CH4, N=5, and, since it i
1) Position Matrix Elements of the Harmonic Oscillator
Focus first on I1 , i.e.:
1 m 1 / 4
m
I1 = dx
x exp(m x 2 / 2
2
2
=
=
2m m
2m
2m
1/ 2
=
m 1/ 4
2
) x
exp( m x / 2
)
dx exp( m x 2 / )
Chem. 1410: Solutions for Hand-In Problems.
a) The normalization constraint is:
1 = dx * ( x) ( x) =
1/ 2
x2
dx exp( 2 2 )
1
2 2
The integral identity given in part a)-ii) confirms that ( x) in Eq
Chem. 1410: Solutions for Hand-In Problems.
1)(2 points) Consider
xpF ( x) = x
F ( x)
i x
(i)
Now consider:
pxF ( x) =
[ xF ( x)]
=
i
x
i
F ( x)
x x + F ( x)
Subtraction (i)-(ii), we obtain:
(ii)
Chemistry 1410, Hour Exam 1, Solution Key
1) a) Appeal to the 1D Particle in the Box (PinB) model for the single electron energy levels of the pi
electrons in hexatriene. Furthermore: respect the Paul
Oct. 31, 2007
Chem. 1410
Practice Problem for Hour Exam 2, solution
1) Starting from
v eff (r ) =
1
k (r re ) 2 +
2
2
l (l + 1) / 2 r 2 ,
[A1]
The minimum occurs at the position rm such that dv eff (r