How Does NaCl Molecule Transition Influence Photon Frequency?

[skrat]

Homework Statement

A molecule of NaCl has a reflective potential $C/r^n$, where $n=35$. What is the frequency of radiated photon at oscillating transition of the distance between the atoms is $r_0=0.236nm$. And what is the frequency if the transition is rotational?

Homework Equations

$M(Na)=23kg/kmol$ and $M(Cl)=35kg/kmol$.

The Attempt at a Solution

I have no idea what to do here. I am guessing it goes something like

$\frac{C}{r^n}+\hbar \omega (n+1/2)=h\nu$

But even if that is ok, I can't see a way to determine $C$. This has to be something with units $eVm^{35}$ ?

Could anybody please help me a bit?

 

[ehild]

In the molecule, the Na and Cl are charged and attract each other by their Coulomb interaction, but also repel each other by the interaction of their electron clouds. The equilibrium distance between them is r₀, where the potential energy of the system is minimum. You get the unknown C from that condition.

The molecule vibrates about that equilibrium separation. Determine how the net potential changes near the equilibrium to get the force constant of the oscillation.

That is the vibration of two-bodies with respect to each other. . You can reduce it a single-body vibration by using the reduced mass of the NaCL which is μ=M_Na M_Cl/(M_Na + M_Cl)

ehild

 

[skrat]

So the total potential energy of the system consists of Coulumb interaction and $C/r^n$, therefore

$V(r)=-\frac{e^2}{4\pi \epsilon _0 r}+\frac{C}{r^n}$

$V^{'}(r_0)=\frac{e^2}{4\pi \epsilon _0 r^2}-n\frac{C}{r^{n+1}}=0$

Which gives me $C=\frac{e^2r_0^{n-1}}{4\pi \epsilon _0 n}$.

Right?

 

[ehild]

It is OK.

ehild

 

[skrat]

The molecule vibrates about that equilibrium separation. Determine how the net potential changes near the equilibrium to get the force constant of the oscillation.

ehild

Hmmm...

Since Coulumb potential $\propto \frac{1}{r}$ and repel potential $\propto \frac{1}{r^{35}}$ than I would say that potential $C/r^n$ is not significant for greater $r$.

Which would give me $C/r^n+\hbar\omega (n_f-n_i)=h\nu$ if $f$ stands for final state and $i$ for initial, where $\nu$ is of course frequency of light.

 

[ehild]

The repelling force just balances the attracting Coulomb force at equilibrium. You cannot ignore any of them.
The atoms oscillate about their equilibrium separation with frequency ν. You need to find the potential function near the equilibrium in order to find the frequency of the SHM, and the energy states of the oscillator. Write the Tailor series of the potential in terms of the interatomic distance up to the second order term. The frequency of the emitted photons will be proportional to the vibration frequency of the NaCL molecule.
See http://en.wikipedia.org/wiki/Molecular_vibration , scroll down to "Newtonian Mechanics"ehild

 

[skrat]

1. SHM stands for?
2. Why SECOND order? Is that because harmonic potential is $\propto r^2$?

 

[ehild]

SHM stands for the change of interatomic distance, Δr. The harmonic potential is second order in Δr.

ehild

 

[skrat]

So total potential is than $V(r)=-\frac{e^2}{4\pi \epsilon _0 r}+\frac{C}{r^n}=-\frac{e^2}{4\pi \epsilon _0 r}+\frac{e^2}{4\pi \epsilon _0n r}=\frac{e^2}{4\pi \epsilon _0r}(\frac{1}{n}-1)$

And finally $V(r)=\frac{e^2}{4\pi \epsilon _0r}\frac{1-n}{n}$

$V^{'}(r)=-\frac{e^2}{4\pi \epsilon _0r^2}\frac{1-n}{n}$ and

$V^{''}(r)=\frac{e^2}{2\pi \epsilon _0r^3}\frac{1-n}{n}$

Now if we take a look at Taylor expansion $V(r)=V(r_0)+V^{'}(r_0)(r-r_0)+\frac{1}{2}V^{''}(r_0)(r-r_0)^2+...$ we find out that

$\frac{e^2}{4\pi \epsilon _0r_0^3}\frac{1-n}{n}=\frac{m_r\omega ^2}{2}$ where $m_r$ is reduced mass.

Therefore $\omega=(\frac{e^2}{2\pi \epsilon _0m_rr_0^3}\frac{1-n}{n})^{1/2}$.

I guess now I can write $\hbar \omega ((n_f+1/2)-(n_i+1/2))=h\nu$ and calculate the frequency of the light.

 

[ehild]

So total potential is than $V(r)=-\frac{e^2}{4\pi \epsilon _0 r}+\frac{C}{r^n}=-\frac{e^2}{4\pi \epsilon _0 r}+\frac{e^2}{4\pi \epsilon _0n r}=\frac{e^2}{4\pi \epsilon _0r}(\frac{1}{n}-1)$

That is not correct.

C is constant, it contains r₀ instead of r.

ehild

 

[skrat]

C is constant, it contains r₀ instead of r.

ehild

Huh, what a mistake-a to make-a. :D

$V(r)=-\frac{e^2}{4\pi \epsilon _0 r}+\frac{e^2r_0^{n-1}}{4\pi \epsilon _0 n}\frac{1}{r^n}$

$V(r)=\frac{e^2}{4\pi \epsilon _0}(\frac{r_0^{n-1}}{n}\frac{1}{r^n}-\frac{1}{r})$

$V^{'}(r)=\frac{e^2}{4\pi \epsilon _0}(-\frac{r_0^{n-1}}{r^{n+1}}+\frac{1}{r^2})$

$V^{''}(r)=\frac{e^2}{4\pi \epsilon _0}((n+1)r_0^{n-1}\frac{1}{r^{n+2}}-2\frac{1}{r^3})$

Now using Taylor expansion:

$\frac{m_r\omega ^2}{2}=\frac{1}{2}V^{''}(r_0)=\frac{1}{2}\frac{e^2}{4\pi \epsilon _0}((n+1)\frac{1}{r_0^3}-2\frac{1}{r_0^3})$

Which gives me $\omega=(\frac{e^2(n-1)}{4\pi \epsilon _0r_0^3m_r})^{1/2}$

 

[ehild]

Huh, what a mistake-a to make-a. :D

$V(r)=-\frac{e^2}{4\pi \epsilon _0 r}+\frac{e^2r_0^{n-1}}{4\pi \epsilon _0 n}\frac{1}{r^n}$

$V(r)=\frac{e^2}{4\pi \epsilon _0}(\frac{r_0^{n-1}}{n}\frac{1}{r^n}-\frac{1}{r})$

$V^{'}(r)=\frac{e^2}{4\pi \epsilon _0}(-\frac{r_0^{n-1}}{r^{n+1}}+\frac{1}{r^2})$

$V^{''}(r)=\frac{e^2}{4\pi \epsilon _0}((n+1)r_0^{n-1}\frac{1}{r^{n+2}}-2\frac{1}{r^3})$

Now using Taylor expansion:

$\frac{m_r\omega ^2}{2}=\frac{1}{2}V^{''}(r_0)=\frac{1}{2}\frac{e^2}{4\pi \epsilon _0}((n+1)\frac{1}{r_0^3}-2\frac{1}{r_0^3})$

Which gives me $\omega=(\frac{e^2(n-1)}{4\pi \epsilon _0r_0^3m_r})^{1/2}$

It looks correct. Evaluate.

ehild

 

[skrat]

$\hbar \omega ((n_i+1/2)-(n_f+1/2))=h\nu$

$\frac{h}{2\pi }\omega (n_i-n_f)=h\nu$

and finally

$\nu =\frac{\omega }{2\pi }(n_i-n_f)$ where $\omega=(\frac{e^2(n-1)}{4\pi \epsilon _0r_0^3m_r})^{1/2}=(\frac{34e^2}{4\pi \epsilon _0r_0^3m_r})^{1/2}$.

Therefore $\nu =25.7 THz$ if $\Delta n=1$.

Similary for rotational transitions:

$\Delta E=h\nu$

$\frac{\hbar ^2}{2mr^2}(l_i(l_i+1)-l_f(l_f+1))=h\nu$

$\nu =\frac{h}{8\pi ^2mr_0^2}(l_i(l_i+1)-l_f(l_f+1))$

Which gives me $\nu =13.1 GHz$ if $l_i=1$ and $l_f=0$.If I am not mistaken, this should be it.

Thanks for all the help, ehlid!

ps: I am still confused how is this problem part of introductory physics but, nevermind ...