Ratio of Wave Numbers for Balmer and Brackett Series Spectral Lines

To find the ratio of the wave numbers for the first line of the Balmer series and the first line of the Brackett series for the Hydrogen atomic spectrum, we use the Rydberg formula for wave number ($\bar{\nu}$):

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

For hydrogen atom, $Z = 1$, so: $\bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

1. Wave number of the first line of the Balmer series:

For the Balmer series, the lower energy level is $n_1 = 2$. The first line (lowest energy line) corresponds to the transition from $n_2 = 3$ to $n_1 = 2$:

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

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2. Wave number of the first line of the Brackett series:

For the Brackett series, the lower energy level is $n_1 = 4$. The first line corresponds to the transition from $n_2 = 5$ to $n_1 = 4$:

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

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3. Ratio of the wave numbers:

$\frac{\bar{\nu}_{\text{Balmer}}}{\bar{\nu}_{\text{Brackett}}} = \frac{R_H \left( \frac{5}{36} \right)}{R_H \left( \frac{9}{400} \right)} = \frac{5}{36} \times \frac{400}{9} = \frac{500}{81}$

Expressing the ratio in the form $5 : x$:

$\frac{500}{81} = \frac{5}{x} \implies x = \frac{81}{100} = 0.81$

Thus, the ratio is $5 : 0.81$.

Correct Answer: Option B ($5 : 0.81$)