Question

When a photon is emitted from an atom, the atom recoils. The kinetic energy of recoil and the energy of the photon come from the difference in energies between the states involved in the transition. Suppose, a hydrogen atom changes its state from n = 3 to n = 2. Calculate the fractional change in the wavelength of light emitted, due to the recoil.

Answer 1

Difference in energy in the transition from n = 3 to n = 2 is 1.89 eV ( = E).

If all this energy is used up in emitting a photon (i.e. recoil energy is zero).
Then,
$$\begin{gathered} E=\frac{hc}{\lambda } \\ \Rightarrow \lambda =\frac{hc}{E}...\left(\mathrm{i}\right) \end{gathered}$$

If difference of energy is used up in emitting a photon and recoil of atom, then let E_R be the recoil energy of atom.
$$\begin{gathered} E=\frac{hc}{\lambda \text{'}}+E_R \\ \Rightarrow \lambda \text{'}=\frac{hc}{E-E_R}...\left(\mathrm{ii}\right) \end{gathered}$$
Fractional change in the wavelength is given as,
$$\begin{gathered} \frac{∆\lambda }{\lambda }=\frac{\lambda \text{'}-\lambda }{\lambda } \\ \Rightarrow \frac{∆\lambda }{\lambda }=\frac{1}{\lambda }\left(\frac{hc}{E-E_R}-\frac{hc}{E}\right) \\ \Rightarrow \frac{∆\lambda }{\lambda }=\frac{E}{hc}\left(\frac{hc{E}_R}{E(E-E_R)}\right)\left(\because \lambda =\frac{hc}{E}\right) \\ \Rightarrow \frac{∆\lambda }{\lambda }=\left(\frac{E_R}{(E-E_R}\right) \end{gathered}$$