QJEEM23Phf54c: JEE Main 2023 Physics - Class 12 JEEM23Phf54c Solution: If 917 A o be the lowest wavelength of Lyman series then the…

The energy difference formula for transitions between energy levels in a hydrogen atom, which is given by

$\Delta E = -13.6 \, \text{eV} \times \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

where ( $n_1$ ) and ( $n_2$ ) are the initial and final energy levels, respectively. For the Lyman series, the electron transitions to the ground state (( $n_1$ = 1)), and for the Balmer series, the electron transitions to the first excited state (( $n_1$ = 2)).

From the given information, we have the lowest wavelength in the Lyman series, ( $\lambda_1 = 917 \, \text{Å}$ ). Therefore, the energy difference ( $\Delta E$ ) for the Lyman series is

$\Delta E = \frac{hc}{\lambda_1}$

where (h) is Planck's constant and (c) is the speed of light.

Similarly, for the Balmer series, the energy difference ( $\Delta E$ ) is

$\Delta E = -13.6 \, \text{eV} \times \left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) = -13.6 \, \text{eV} \times \frac{1}{4}$

The corresponding wavelength ( $\lambda_2$ ) is

$\lambda_2 = \frac{hc}{\Delta E}$

By comparing ( $\Delta E$ ) for the Lyman and Balmer series, you found that

$\frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} = \frac{1}{4}$

Therefore, the lowest wavelength in the Balmer series is

$\lambda_2 = 4 \lambda_1 = 4 \times 917 \, \text{Å} = 3668 \, \text{Å}$