# Vibrating molecule of IBr - energies

1. Apr 17, 2014

### skrat

1. The problem statement, all variables and given/known data
Effective potential of atoms in molecule of IBr can be described as $V(r)=V_0[(\frac{r}{a})^{-8}-10(\frac{r}{a})^{-4}]$, where $a=1nm$ and $V_0=0.1eV$. Calculate the first three vibration states if the potential close to minimum is harmonic. $M(I)=127g/mol$ and $m(Br)=80g/mol$.

2. Relevant equations

3. The attempt at a solution

$V(r)=V_0[(\frac{r}{a})^{-8}-10(\frac{r}{a})^{-4}]$

$V^{'}(r)=V_0[-8\frac{r^{-9}}{a^{-8}}+40\frac{r^{-5}}{a^{-4}}]$

which gives me $r_0=(\frac{8}{40})^{1/4}a=0.669a$ and

$V^{''}(r)=V_0[72\frac{r^{-10}}{a^{-8}}-200\frac{r^{-6}}{a^{-4}}]$

Now looking at Taylor expansion $V(r)=V(r_0)+V^{'}(r_0)(r-r_0)+\frac{1}{2}V^{''}(r_0)(r-r_0)^2+...$

$\frac{m_r\omega ^2}{2}=\frac{1}{2}V_0[72\frac{r_0^{-10}}{a^{-8}}-200\frac{r_0^{-6}}{a^{-4}}]$

$m_r\omega ^2=V_0[72\frac{a^8}{r_0^10}-200\frac{a^4}{r_0^6}]=V_0[72\frac{1}{(0.669a)^2}-200\frac{1}{(0.669a)^2}]$

$m_r\omega ^2=\frac{V_0}{(0,669a)^2}[72-200]$ where $m_r=\frac{m(I)m(Br)}{m(I)+m(Br)}=49.08u$.

Finally

$\omega =(\frac{V_0}{m_r(0.669a)^2}[27-200])^{1/2}=7,5\cdot 10^{12}Hz$

Now $E_n=\hbar \omega (n+1/2)=0.0049(n+1/2)$ and $E_0=0.0023eV$.

Now why is that wrong? Should I also take Coulomb potential into account?

Last edited: Apr 17, 2014
2. Apr 17, 2014

### ehild

The whole potential function is given, no need to add anything.
The problem asks the first three vibration states. I think, you need to give the energies of these states, not only one energy.

ehild

3. Apr 17, 2014

### skrat

$E_0=0.0023eV$,
$E_1=0.0074eV$ and
$E_2=0.0123eV$.

Thanks for checking. I wasn't sure everything is ok.