What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

The wave number $\bar{\nu}$ of a spectral line in the hydrogen emission spectrum is given by the Rydberg formula:

$\bar{\nu} = R_H \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$

For the first line (lowest energy) of the Balmer series, the transition is from $n_2 = 3$ to $n_1 = 2$:

$\bar{\nu}_{Balmer} = R_H \left( \dfrac{1}{2^2} - \dfrac{1}{3^2} \right) = R_H \left( \dfrac{1}{4} - \dfrac{1}{9} \right) = \dfrac{5R_H}{36}$

For the first line of the Brackett series, the transition is from $n_2 = 5$ to $n_1 = 4$:

$\bar{\nu}_{Brackett} = R_H \left( \dfrac{1}{4^2} - \dfrac{1}{5^2} \right) = R_H \left( \dfrac{1}{16} - \dfrac{1}{25} \right) = \dfrac{9R_H}{400}$

Taking the ratio of the two wave numbers:

$\dfrac{\bar{\nu}_{Balmer}}{\bar{\nu}_{Brackett}} = \dfrac{\dfrac{5R_H}{36}}{\dfrac{9R_H}{400}} = \dfrac{5}{36} \times \dfrac{400}{9} = \dfrac{500}{81}$

To express this ratio in the form $5 : x$, we divide the numerator and the denominator by $100$:

$\dfrac{500}{81} = \dfrac{5}{0.81}$

The ratio is $5 : 0.81$.

Answer: $5:0.81$