Vibrating molecule of IBr – energies

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?

http://ift.tt/1pcum1t

Leave a comment

Your email address will not be published.


*


Show Buttons
Hide Buttons