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:

$$ p = \frac{h}{\lambda} $$

From the Rydberg formula

$$ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) $$

$$ p \propto \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) $$

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

For the 1st line of the Balmer series, the electron transitions from $$ n_i = 3 $$ to $$ n_f = 2 $$:

$$ p_1 \propto \left( \frac{1}{2^2} - \frac{1}{3^2} \right) $$

$$ p_1 \propto \left( \frac{1}{4} - \frac{1}{9} \right) $$

$$ p_1 \propto \frac{9 - 4}{36} $$

$$ p_1 \propto \frac{5}{36} $$

For the 2nd line of the Balmer series, the electron transitions from $$ n_i = 4 $$ to $$ n_f = 2 $$:

$$ p_2 \propto \left( \frac{1}{2^2} - \frac{1}{4^2} \right) $$

$$ p_2 \propto \left( \frac{1}{4} - \frac{1}{16} \right) $$

$$ p_2 \propto \frac{4 - 1}{16} $$

$$ p_2 \propto \frac{3}{16} $$

Now, we find the ratio of their momentums:

$$ \text{Ratio} = \frac{p_1}{p_2} $$

$$ \text{Ratio} = \frac{5/36}{3/16} $$

$$ \text{Ratio} = \frac{5}{36} \times \frac{16}{3} $$

$$ \text{Ratio} = \frac{5 \times 4}{9 \times 3} $$

$$ \text{Ratio} = \frac{20}{27} $$

Comparing this with the given ratio $$ \frac{\alpha}{\beta} $$, the possible values are:

$$ \alpha = 20 $$

$$ \beta = 27 $$