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 (rm ) / dr = 0 . Thus the equation we need
to solve is:
0
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 shown in Fig. 1.
1
delEp( R)0.5
0
0
0
2
4
6
R
Fig. 1. delE
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: ij = 1 if i = j and ij = 0 if i j .] Thus, the overlap
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 =
, 0, .
l (l + 1) = 2 , and the allowed (measurable) val
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 is a nonlinear molecule, nv = 3N-6 = 9.
(1) From class n
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 factor) exp(ik ( E ) x), exp(ik ( E ) x) ; or,
equivalently,
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 / )x 2
2m
[Note: to go from 2nd to 3rd lines, use the give
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. [1] is properly normalized.
b) [Again ] When a quantu
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)
[ xp px]F ( x) = i F ( x) , QED.
2) a)(2 points) The G
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 Pauli Exclusion Principle, putting no more than two
(spin-p
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 / d ) = -(c/ 2 ) . [If
frequency increases, wavelength