Ratio of Momentum of Photons in Balmer Series

The momentum $p$ of a photon is directly proportional to its energy $E$ (or wavenumber $\frac{1}{\lambda}$): $p = \frac{E}{c} = \frac{h}{\lambda}$

For a hydrogen atom, the wavenumber for a transition from an upper energy level $n_2$ to a lower energy level $n_1$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ where $R$ is the Rydberg constant.

For the Balmer series, the lower energy level is $n_1 = 2$.

1. For the $1^{\text{st}}$ line of the Balmer series ($n_2 = 3 \to n_1 = 2$): $\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5}{36}R$ Therefore, the momentum of the photon emitted in the $1^{\text{st}}$ line is: $p_1 = h R \left( \frac{5}{36} \right)$

2. For the $2^{\text{nd}}$ line of the Balmer series ($n_2 = 4 \to n_1 = 2$): $\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3}{16}R$ Therefore, the momentum of the photon emitted in the $2^{\text{nd}}$ line is: $p_2 = h R \left( \frac{3}{16} \right)$

3. Ratio of the momenta: $\frac{p_1}{p_2} = \frac{\frac{1}{\lambda_1}}{\frac{1}{\lambda_2}} = \frac{\frac{5}{36}}{\frac{3}{16}} = \frac{5}{36} \times \frac{16}{3} = \frac{20}{27}$

Given that the ratio of the momentum is $\frac{\alpha}{\beta}$: $\frac{\alpha}{\beta} = \frac{20}{27}$

Thus, the possible values of $\alpha$ and $\beta$ are $20$ and $27$ respectively.

Correct Option: D (20 and 27)