QJEEM24Ph17c1: JEE Main 2024 Physics - Class 12 JEEM24Ph17c1 Solution: The ratio of the shortest wavelength of Balmer series to the…

The wavelength of light emitted when an electron transitions between energy levels in a hydrogen atom is given by the Rydberg formula:

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

where:

• $\lambda$ is the wavelength of the emitted light,
• $R$ is the Rydberg constant,
• $n_1$ is the lower energy level, and
• $n_2$ is the higher energy level.

The Balmer series corresponds to electron transitions where the lower energy level, $n_1$ , is 2, and the Lyman series corresponds to transitions where $n_1$ is 1. The shortest wavelength in each series occurs for transitions from the highest possible energy level ( $n_2 = \infty$ ) to the specified lower level ( $n_1$ ).

For the Balmer series (shortest wavelength):

$n_1 = 2, n_2 = \infty$ ,

Putting these values into the Rydberg formula gives:

$\frac{1}{\lambda_{B}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$

$\frac{1}{\lambda_{B}} = R \left( \frac{1}{4} \right)$

$\lambda_{B} = \frac{1}{R \cdot \frac{1}{4}} = \frac{4}{R}$

For the Lyman series (shortest wavelength):

$n_1 = 1, n_2 = \infty$ ,

Putting these values into the Rydberg formula gives:

$\frac{1}{\lambda_{L}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)$

$\frac{1}{\lambda_{L}} = R \cdot 1$

$\lambda_{L} = \frac{1}{R}$

The ratio of the shortest wavelength of Balmer series ( $\lambda_{B}$ ) to the shortest wavelength of Lyman series ( $\lambda_{L}$ ) is:

$\frac{\lambda_{B}}{\lambda_{L}} = \frac{\frac{4}{R}}{\frac{1}{R}} = \frac{4}{1} = 4:1$

Therefore, the correct answer is Option D: $4: 1$ .