# Homework Help: Wave Function_vibtrating nuclei

1. Oct 2, 2014

### terp.asessed

1. The problem statement, all variables and given/known data
The Ground state wave function governing the motion of a pair of vibrating nuclei looks like:

wave function = wave (x) = (a/pi)1/4 e-ax2/2

where a = alpha = mu*w/(h bar), which is determined by mu = reduced mass of the pair
where w = angular vibrational frequency

...So, suppose the vibrating atoms in question were C and O. Compute what v~, the frequency of the vibration in cm-1, would have to be if the root-mean square fluctation in bond length were 0.035Angstom.

I lost track of the units for my solution, and I am not sure if I have done correctly...

2. Relevant equations
wave function = wave (x) = (a/pi)1/4 e-ax2/2

a = alpha = mu*w/(h bar), which is determined by mu = reduced mass of the pair

w = angular vibrational frequency

root mean square = <x2> - <x>2

3. The attempt at a solution
root mean square = <x2> - <x>2

where <x2> = integral (-infinite to infinite) wave(x) x2 wave(x) dx
= which I got 1/(2a) as a result

<x> = integral (x=o to L) p(x) x p(x) dx which I got as 0 in the end

so...I input:
root mean square = <x2> - <x>2 = 1/(2a) - 0 = 1/(2a)

since
root mean square = 0.035 angstrom = 1/(2a) and b/c a = mu*w/(h bar), I rearranged so that:

w = (h bar)/(0.07mu)...where mu = m1m2/(m1 + m2)

For mass of 1 C atom: 12.01g/(6.022x1023 atoms) x 1kg/103g = 1.99 x 10-26kg

For mass of 1 O atom: 16.00g/(6.022x1023 atoms) x 1kg/103g = 2.66 x 10-26kg

mu = 1.14 x 10-20kg

therefore w = (h/(2pi))1/(0.07mu) = 1.32 x 10-13 omega (is this right????) I know h has units of J*2 and mu = kg.....but together they do NOT make omega.....

therefore v~ = w/(2pi*c) where c = speed of light = 1.32 x 10-13 (assuming the unit is correct) / (2pi*3x108m/s) = 7.02 x 10-23 s-1

b/c I got lost track of units....I am not sure if the units in the solution is right...if someone could (1) explain the units in each calculations I did and then (2) give me hints as how to keep track of units for near future too, that would be so wonderful.

2. Oct 2, 2014

### TeethWhitener

You made a numerical error in your value of $\mu$. If your mass for a carbon atom is ~10-26kg and your mass for an oxygen atom is ~10-26kg, there's no way the reduced mass will be ~10-20kg. Fix that first. As for the units, Planck's constant $\hbar$ is in J s (joule-seconds). Maybe breaking this (and everything else) down into base units (kg, m, s) will help. Just a heads up, you also forgot the "root" part of "root mean square."

3. Oct 2, 2014

### terp.asessed

Hello--thanks you for the reply--so, you're right, I miscalculated: μ = 1.14 x 10-26 kg

Also, since,
ℏ = J*s = kg*m2/s2....is omega a right unit for w?

w = (h/(2pi))1/(0.07mu) = 1.32 x 10-7 (J*s)/kg = 1.32 x 10-7 m2/s....???? Did I make mistake? It is of acceleration unit, not velocity...
Aside, could you please clarify by what you mean by "root part of the root mean square"? If you could, thanks!

4. Oct 2, 2014

### TeethWhitener

You're close with $\hbar$. Joules are kg*m2/s2 (you can remember this by recalling that joules are a measure of energy and kinetic energy is $\frac{1}{2}mv^2$), so J*s is kg*m2/s. Also, it's clear that you know that $\mu$ has units of kg, but 0.07 also has units. This comes from the root mean square of distance, which, by the way, is $$\sqrt{\langle x^2 \rangle - \langle x \rangle ^2 }$$ not just $$\langle x^2 \rangle - \langle x \rangle ^2$$ The units of rms distance are the same as the units of plain old distance, which in this case is just angstroms. Hopefully that should get you started.