JEE Physics 2023 Question with Solution

Given: The lowest wavelength of Lyman series is $917 \, \text{Å}$.

Find: The lowest wavelength of Balmer series.

For the Lyman series, the lowest wavelength corresponds to the transition from $n = \infty$ to $n = 1$.

For the Balmer series, the lowest wavelength corresponds to the transition from $n = \infty$ to $n = 2$.

The energy $E_0$ for the Lyman series is related to the wavelength $\lambda_0$ by:

$$E_0 = \frac{hc}{\lambda_0}$$

For Lyman, $\lambda_0 = 917 \, \text{Å}$, so:

$$E_0 = \frac{hc}{917 \, \text{Å}}$$

For the Balmer series, the energy is related by:

$$\frac{E_0}{4} = \frac{hc}{\lambda}$$

Substituting the energy of the Lyman series:

$$\frac{hc}{4 \times 917 \, \text{Å}} = \frac{hc}{\lambda}$$

Thus, the wavelength for the Balmer series is:

$$\lambda = 917 \times 4 = 3668 \, \text{Å}$$

Therefore, the lowest wavelength of the Balmer series is $3668 \, \text{Å}$.

Given: The series-limit wavelength of Lyman series is $917 \, \text{Å}$.

Find: The series-limit wavelength of Balmer series.

At the series limit, the emitted photon energy for transition to $n = 1$ is proportional to $1$, while for transition to $n = 2$ it is proportional to $\frac{1}{4}$.

Hence, the Balmer-limit photon has one-fourth the energy of the Lyman-limit photon. Since

$$E = \frac{hc}{\lambda}$$

wavelength is inversely proportional to energy.

So, if energy becomes one-fourth, wavelength becomes four times:

$$\lambda_{\text{Balmer}} = 4\lambda_{\text{Lyman}}$$

Substitute the given value:

$$\lambda_{\text{Balmer}} = 4 \times 917 = 3668 \, \text{Å}$$

Therefore, the required numerical value is 3668.