First, let us identify the transitions for the first members of the Balmer and Paschen series in the hydrogen atom:

Let wavelengths be:

The wavelength for a transition in the hydrogen atom is given by the Rydberg formula:

$$ \frac{1}{\lambda} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) $$

For the first Balmer line $(3 \to 2)$:

$$ \frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) $$

Calculate:

$$ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} $$

So,

$$ \frac{1}{\lambda_1} = R_H \cdot \frac{5}{36} $$

For the first Paschen line $(4 \to 3)$:

$$ \frac{1}{\lambda_2} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{9} - \frac{1}{16} \right) $$

Calculate:

$$ \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} $$

So,

$$ \frac{1}{\lambda_2} = R_H \cdot \frac{7}{144} $$

The frequency is given by $f = \dfrac{c}{\lambda}$

Therefore,

$$ f_1 = \frac{c}{\lambda_1}, \quad f_2 = \frac{c}{\lambda_2} $$

$$ \frac{f_1}{f_2} = \frac{\lambda_2}{\lambda_1} $$

From above:

$$ \frac{1}{\lambda_1} = R_H \cdot \frac{5}{36} \implies \lambda_1 = \frac{36}{5 R_H} $$

$$ \frac{1}{\lambda_2} = R_H \cdot \frac{7}{144} \implies \lambda_2 = \frac{144}{7 R_H} $$

Therefore,

$ \frac{f_1}{f_2} = \frac{\lambda_2}{\lambda_1} = \frac{\frac{144}{7 R_H}}{\frac{36}{5 R_H}} = \frac{144}{7} \times \frac{5}{36} $

$ = \frac{144 \times 5}{7 \times 36} $

$ = \frac{720}{252} $

$ = \frac{720 \div 36}{252 \div 36} = \frac{20}{7} $

Correct answer:

$ \boxed{20:7} $

So, the answer is Option A.

Atoms And Nuclei

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