JEE Physics 2025 Question with Solution

Given: A hydrogen atom is considered, and we need the ratio of the largest wavelength of the Lyman series to that of the Balmer series.

Find: $\dfrac{\lambda_{\text{Lyman}}}{\lambda_{\text{Balmer}}}$

Use the Rydberg formula:

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

For the Lyman series, transitions end at $n_1 = 1$. For the Balmer series, transitions end at $n_1 = 2$.

The largest wavelength in a series corresponds to the smallest energy gap, so we take the first line of each series.

For the Lyman series, the largest wavelength corresponds to the transition $n_2 = 2 \to n_1 = 1$:

$$\frac{1}{\lambda_{\text{Lyman}}} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = \frac{3R_H}{4}$$

Therefore,

$$\lambda_{\text{Lyman}} = \frac{4}{3R_H}$$

For the Balmer series, the largest wavelength corresponds to the transition $n_2 = 3 \to n_1 = 2$:

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

Therefore,

$$\lambda_{\text{Balmer}} = \frac{36}{5R_H}$$

Now take the ratio:

$$\frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Balmer}}} = \frac{\frac{4}{3R_H}}{\frac{36}{5R_H}} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27}$$

Therefore, the correct option is B.