The ratio of momentum of the photons of the 1^ st and 2^ nd line of...

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 $27$