The ratio of momentum of the photons of the \(1^{st}\) and \(2^{nd}\) line of Balmer series of Hydrogen atoms is \(\alpha/\beta\). The possible values of \(\alpha\) and \(\beta\) are:-

The momentum of a photon is given by \(p = \dfrac{h}{\lambda}\).

From Rydberg's formula, the wave number is \(\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)\).

For the first line of the Balmer series, the transition is from \(n_2 = 3\) to \(n_1 = 2\).

\(p_1 = \dfrac{h}{\lambda_1} = hR \left( \dfrac{1}{2^2} - \dfrac{1}{3^2} \right) = hR \left( \dfrac{1}{4} - \dfrac{1}{9} \right) = \dfrac{5hR}{36}\)

For the second line of the Balmer series, the transition is from \(n_2 = 4\) to \(n_1 = 2\).

\(p_2 = \dfrac{h}{\lambda_2} = hR \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right) = hR \left( \dfrac{1}{4} - \dfrac{1}{16} \right) = \dfrac{3hR}{16}\)

The ratio of the momenta is:

\(\dfrac{p_1}{p_2} = \dfrac{\dfrac{5hR}{36}}{\dfrac{3hR}{16}} = \dfrac{5}{36} \times \dfrac{16}{3} = \dfrac{20}{27}\)

Comparing with \(\dfrac{\alpha}{\beta}\), the possible values are \(\alpha = 20\) and \(\beta = 27\).

Answer: \(20\) and \(27\)