Wave Function_vibtrating nuclei

[terp.asessed]

Homework Statement

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...

Homework 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

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.

 

[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."

 

[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!

 

[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.