Question

Starting with equation (10-4), show that if $\mathbf{\Delta n}$ is$\mathbf{-}\mathbf{1}$ as a photon is emitted by a diatomic molecule in a transition among rotation-vibration states, but $\mathbf{\Delta \ell }$can be$\mathbf{\pm }\mathbf{1}$ . Then the allowed photon energies obey equation (10-6).

Short answer

The photon’s energy is equal to${\mathrm{E}}_{\mathrm{photon}}=\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}\pm 1\frac{{\mathrm{h}}^2}{{\mathrm{\mu a}}^2}$ .

Step-by-step solution

Calculate the energy change for a molecule

Let us calculate the energy change for a molecule changing from state n,l to state n-1, l+1, this is the photons energy.

$\begin{array}{c}{\mathrm{E}}_{\mathrm{nl}}-{\mathrm{E}}_{\mathrm{n}-\mathrm{l},\mathrm{l}+1}=\left(\mathrm{n}+\frac{1}{2}\right)\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}+\frac{{\mathrm{h}}^2\times \mathrm{l}\times (\mathrm{l}+1)}{2{\mathrm{\mu a}}^2}-\left(\left(\mathrm{n}-1+\frac{1}{2}\right)\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}+\frac{{\mathrm{h}}^2\times (\mathrm{l}+1)\times (\mathrm{l}+2)}{2{\mathrm{\mu a}}^2}\right)\\ =\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}+\frac{{\mathrm{h}}^2\times (\mathrm{l}-(1+2\left)\right)\times (\mathrm{l}+1)}{2{\mathrm{\mu a}}^2}\\ =\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}-\frac{{\mathrm{h}}^2\times (\mathrm{l}+1)}{{\mathrm{\mu a}}^2}\end{array}$

Then let us calculate the energy change for a molecule changing from state n,l to state n-1,l-1. This is the photon’s energy.

$\begin{array}{c}{\mathrm{E}}_{\mathrm{nl}}-{\mathrm{E}}_{\mathrm{n}-\mathrm{l},\mathrm{l}-1}=\left(\mathrm{n}+\frac{1}{2}\right)\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}+\frac{{\mathrm{h}}^2\times \mathrm{l}\times (\mathrm{l}+1)}{2{\mathrm{\mu a}}^2}-\left(\left(\mathrm{n}-1+\frac{1}{2}\right)\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}+\frac{{\mathrm{h}}^2\times (\mathrm{l}-1)\times \left(\mathrm{l}\right)}{2{\mathrm{\mu a}}^2}\right)\\ =\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}+\frac{{\mathrm{h}}^2\times (\mathrm{l}+1-(\mathrm{l}-1\left)\right)\times \left(1\right)}{2{\mathrm{\mu a}}^2}\\ =\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}-\frac{{\mathrm{h}}^2\times \mathrm{l}}{{\mathrm{\mu a}}^2}\end{array}$

Conclusion.

Thus the photon’s energy is equal to

$\begin{array}{c}{\mathrm{E}}_{\mathrm{photon}}=\mathrm{h}\sqrt{\frac{\mathrm{\kappa }}{\mathrm{\mu }}}\pm 1\frac{{\mathrm{h}}^2}{{\mathrm{\mu a}}^2}\\ \mathrm{I}=1,2,3,\mathrm{..........}\end{array}$