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Chemistry 163A - Quantum Mechanics and Spectroscopy

Question and Answer Page - Fall 2002

Send questions regarding course material via e-mail to Professor Switkes. Look for a response to be posted on this page! e-mail does not sleep!

Date Topics
Mon, 30 Sep 2002 21:07:57 HW #2, problem #5
TTue, 08 Oct 2002 09:25:2
Wed, 9 Oct 2002 07:20:22		 HW 3 Question 13; HW #13
Tue, 08 Oct 2002 09:25:29 HW 3 Question 13,15
Mon, 14 Oct 2002 15:41:57 HW Prob #19
Mon, 14 Oct 2002 23:00:34 HW# Prob #17
Mon, 21 Oct 2002 16:06:5 HW# Prob #21

DATE:

Tue, 1 Oct 2002 08:53:09


QUESTION: HW #2 Problem #5


Chemistry 163A Student Asks:
Gene,

I understand for problem 5 a. you substitute d^2 x/dt^2 and by doing so
you get the differential equation for time with respect to distance.
However, I was trying to figure out how take the integral of the
function to get a function of X(t) such that time is variable and yields
a particular position. As the function stands, it is only dependent on
position such that:
(d^2 x/dt^2 ) = (-k x) / m
When I attempted taking to solve the equation I was struggling to recall
how to integrate this in a fashion that will yield a first derivative.
I believe I am attempting to have both t and x as variables, but when I
do this I get:
from:
d^2 x/x = - (k/m) dt^2
I integrated this to:
(ln x)dx = -(k/m) t dt
Yet when I try to find a way to use this to solve for t = 0 and position=A,
clearly I lose one side of the equation due to the zero. I'm
currently at a loss as to where my approach is wrong. Could you please
help to point out any flaws in my approach taht you might see?


Thanks,
[student's name]

ANSWER:



Hi [student name],


First, note that problem 5a didn't call for acutally obtaining
the solution to the differential equation;
5a required just setting up the equation; part 5c
is a "show that" given the solution. In 5b, the question
is just to give the value of the ?, ie the velocity
of the fully extended spring, one of the boundary conditions.

However, since you are interested in the full solution:
The solution of the classical harmonic oscillator requires
some substitutions; directly integrating d^2 is not usually
profitable; d^2x is not (dx)^2. THE BOTTOM LINE IS THAT
YOU CANNOT GENERALLY SOLVE SECOND ORDER D.E.'S BY SIMPLE
INTERGRATION WITHOUT FURTHER TRICKS.

Example 2-4 on page 53 of McQ is the solution to the harmonic
oscillator with w^2=k/m. Following the development on pp 50-53,
you can see that, in general, solving differential equations
requires 'tricks' [substitution of exp(alpha*x)].

Another way to solve (d^2 x/dt^2 ) = -(k/m) x would be
to make the substitution f(x)= dx/dt [another d.e.TRICK!!].
Then (d^2 x/dt^2 )=df/dt=dx/dt df/dx (chain rule)= f df/dx.

The equation for f(x) is:
f df/dx= - (k/m) x which can be solved by integration:
f^2= -(k/m) x^2
f= + or - (-1)^1/2 (k/m)^1/2 x
dx/dt= +- i (k/m)^1/2 x
the above again can be solved by simple integration

dx/x =+- i (k/m)^1/2 dt
ln x- = +- i (k/m)^1/2 t
x= exp [i (k/m)^1/2 t] or x= exp [-i (k/m)^1/2 t]

the general solution would be a combination of these two complex
exponentials; and the boundary conditions can be used to evaluate:
x(t)= C exp[i (k/m)^1/2 t] + C' exp [-i (k/m)^1/2 t]

with C= A/2 and C' = A/2
one gets x(t)= A cos [ (k/m)^1/2 t]
which satisfies x(0)=A and (dx/dt) t=0 = 0

A little more than you may have bargained for, but I hope that helps!!

Eugene Switkes
Professor of Chemistry

DATE:

Tue, 1 Oct 2002 08:53:09 P>

QUESTION:


Chemistry 163A Student Asks:
[student's name]

ANSWER:



Hi [student name],



Eugene Switkes
Professor of Chemistry

DATE:

Tue, 8 Oct 2002 09:36:20 and P>

QUESTION: HW 3 Question 13


Two Chemistry 163A Students Ask:
Hi.

* For the table you want us to set up you have absorption values you
want us to calculate and compare to observed values, V/ cm. I
cannot figure out how to calculate my own values for this column.
Is there a formula that I missed?
* In question 15, what do you mean by: "looking"?

Thank you / [student's name]
and


Gene,

On problem number 13, I got the change in energies for
all of the compounds from highest occupied to lowest
unoccupied levels but my answers are all in J. How do
I convert them to (cm^-1)?

Thanks,

[student's name]
Chem 163A

ANSWER:



Hi [student name],


the column (~v)calc refers to the energy of transitions calculated according
to the discussion in part (c) above. One must convert from deltaE (energy of
transisiton) units to cm-1 (another unit frequently used to
specify specrtal 'energies'). The relationship is
deltaE= hv= hc(1/lambda)=hc (~v)
thus (~v)calc = (delta E)calc/hc
or the conversion from MKS energy units to cm-1 is given in the coversion
table on the (front? back?) cover of Mc Quarrie

and
[student's name],

Hope you got your question answered already,
but

1. see front inside cover of McQuarrie for conversion: 1 J= 5.035 x 10^22 cm-1
2. the conversion factor is based on
E=hv=hc/lambda
v~[cm-1]=(1/lambda[cm])=E [J]/ (h[J s] c[ cm s^-1])

Eugene Switkes

Professor of Chemistry

DATE:

Tue, 8 Oct 2002 09:36:20


QUESTION: HW3 Question 15


Chemistry 163A Student Asks:

* In question 15, what do you mean by: "looking"?

ANSWER:


"looking" at the n->infinity behavior of the PIB wavefunction
relates to how does the probability of finding particle
in box change as n->infinity; "where" in box is electron likely
to be found as n->infinity; is the electron more localized
in position (sigmax small) or less localized in position
(sigmax larger) as n-> infinity.



Eugene Switkes
Professor of Chemistry
P>

DATE:

Mon, 14 Oct 2002 15:49:19

QUESTION: HW #19


Chemistry 163A Student Asks:


Hey Gene -

Thanks for the clarification on problem #17, I believe I have that one
under control now.

Problem #19 is only eluding me in that I realized I'm not sure what
value to plug in for k, in order to evalute ~v. If you could simply
answer this question, I believe I'll have the entire problem down.

Thanks again,

[student's name]

ANSWER:



Hi [student name],


Problem #19 does not ask to evaluate ~v; these are given and
are USED to evaluate k (spring constants).

What #19 does ask for is to evaluate the "root mean square (RMS)
displacement " for n=0 ; so
as mentioned in today's class
one needs sigma=[x^2]^1/2 for the n=0 state


Hope that clears it up,

Eugene Switkes
Professor of Chemistry

DATE:

Tue, 15 Oct 2002 05:46:09


QUESTION: HW Problem #17


Chemistry 163A Student Asks:


Hi
* I have a question on #17. I can't get a reasonable
integer for
"n" in part b. The formula I am using is

n^2 = ( E8ml^2 ) / h^2

Is this the right one?

* What does it mean when an energy level is degenerate?

Thanks! / [student's name]

ANSWER:



Hello [student name],


>
>Hi
>
> * I have a question on #17. I can't get a reasonable integer for
> "n" in part b. The formula I am using is
>

Depends on what you mean as "reasonable"?? maybe an "unreasonable" value
tells us something!!
Since this is 3D problem
E(subn)= [nx^2 + ny^2 +nz^2] h^2/(8ml^2)
[nx^2 + ny^2 +nz^2]= E 8ml^2/h^2
and assuming, as stated in problem for E(sub n), nx=ny=nz==n:
3n^2= E 8ml^2/h^2
n^2= E (8ml^2/3h^2)
but the factor of 3 shouldn't change "reasonableness much"

> n^2 = ( E8ml^2 ) / h^2
>
> Is this the right one?
>
As stated in class yesterday-
for part c you will also need E(sub n+1) with nx=n+1, ny=n, nz=n
here E(sub n+1)= [(n+1)^2 + n^2 + n^2] h^2/ (8ml^2)
and delta E/ E = {E(sub n+1)- E(sub n)}/E(sub n);
do the subtraction in the numerator before you put in the numerical value
for n from part b.

>
> * What does it mean when an energy level is
degenerate?
>

A degenerate energy level is one where two independent states
(eigenfunctions of Hop) have the same energies (eigenvalues).
The examples in class were for the 3D particle in a CUBIC box;
here, for example, the states (nx=1,ny=1, nz=2), (nx=1,ny=2, nz=1),
and (nx=2,ny=1, nz=1) all have the same energy level E= 6 h^2/(8 m l^2)
and the level is triply degenerate. When we get to the hydrogen
atom we will find that the px,py, and pz states have the same
energy and are degenerate.

>Thanks! / [student's name]
You are welcome
Hope that helps!!
Eugene Switkes
Professor of Chemistry

DATE:

Mon, 21 Oct 2002

QUESTION:


Gene,

I've been able to derive the d^2/dx^2 for the wave function, and I tried
plugging it into the Schrodinger eqn, but could find no way to further
simplify it. What I have at the moment is:

H (wave function) = (-hbar^2 /2µ) (a / 64pi)^ (1/4) exp^(-c^2 /2)
(4c^2 - 22c + 10) + (1/2)k x^2 (wave function)

The method to which I can get the original function back out of this
situation evades any attempts I've made at it, thus far. I'm sure this
is yet another calculus situation that is just likely to hit me when you
clarify it - but I'm unable to move further at the moment. Any
assistance would be appreciated.

Thanks always,
[student's name]

ANSWER:


Don't forget the value of the constant a (alpha) includes some mus, nu0s (v0),
and hbars which you might want to substitue in
the appropriate places; for example get everything
in terms of mu and k or in terms of alpha and nu0 (v0).
This left-hand side (Hpsi) will eventually have
to equal E psi, where E=E2.


Eugene Switkes
Professor of Chemistry

DATE:


QUESTION:


ANSWER:


Eugene Switkes
Professor of Chemistry