JEE Main 2026 — Structure of Atom Question with Solution

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

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

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

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

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

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

Taking the ratio of the two wave numbers:

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

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

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

The ratio is $5:0.81$.

Answer: $5:0.81$