If ' $\lambda_1$ ' and ' $\lambda_2$ ' are the wavelengths of the first member of the Balmer and Paschen series, in hydorgen atom respectively, then the ratio of respective frequencies, $f_1 / f_2$, is

The ratio of frequencies $f_1/f_2$ for the first members of the Balmer and Paschen series in hydrogen is determined using the Rydberg formula:

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

For the Balmer series ($n_f = 2$), the first member corresponds to $n_i = 3$:

$\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{5R}{36}$

The frequency is $f_1 = c/\lambda_1 = 5Rc/36$.

For the Paschen series ($n_f = 3$), the first member corresponds to $n_i = 4$:

$\frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = \frac{7R}{144}$

The frequency is $f_2 = c/\lambda_2 = 7Rc/144$.

The ratio becomes:

$\frac{f_1}{f_2} = \frac{5Rc/36}{7Rc/144} = \frac{5}{36} \times \frac{144}{7} = \frac{20}{7}$

Final answer: $\boxed{20:7}$