Question

A particular vibrating diatomic molecule may be treated as a simple harmonic oscillator. Show that a transition from that n=2state directly to n=0ground state cannot occur by electric dipole radiation.

Short answer

A transition from n=2 state directly to n=0 ground state cannot occur by electric dipole radiation.

Step-by-step solution

Formula used

The wave function $\psi$represents the wave function of a system and it describes the state of that system. If it is the solution of the time dependent Schrodinger equation, it specifies the sate of a system with a specific energy.

It would be possible only if
$\int r{\psi }_f^{*}\left(\overline{r}\right){\psi }_i\left(\overline{r}\right)dvnotequalto0$

Here, localid="1659722528727" ${\psi }_f$is the final state would be equal to n=0 state and localid="1659722538437" ${\psi }_i$is the initial state would be equal to n = 2 state, r would be x, and dv would be dx, $\psi$is the wave function, and x = r is the separation.

Conclusion:

You also know that,
${\psi }_0={\left(\frac{b}{\sqrt{\pi }}\right)}^{1/2}{e}^{-\frac{1}{2}{b}^2{x}^2}$

Where,$b={(mx/h^2)}^{1/4}$, h is Planck’s constant, m is the mass.

Also,
localid="1659722576928" ${\psi }_2={\left(\frac{b}{8\sqrt{\pi }}\right)}^{1/2}(4b^2{x}^2-2)e^{-\frac{1}{2}{b}^2{x}^2}$

You can clearly see that both the above functions are even functions of x , so the integrand would be odd and the integral from$x=-\infty$to$x=\infty$would be zero.

Hence, a transition from n=2 state directly to n=0 ground state cannot occur by electric dipole radiation.