The shortest wavelength in the Balmer series of hydrogen atom would be

(Given $R= 1.1 × 10^7 m^{-1}$)

3636 Å

The correct answer is Option (1) → 3636 Å

Explanation:

For the hydrogen atom, wavelength in any spectral series is given by the Rydberg formula:

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

where $ R_H = 1.097 \times 10^7\ \text{m}^{-1} $.

For the Balmer series, $ n_1 = 2 $ and $ n_2 = 3, 4, 5, \dots $

The shortest wavelength corresponds to the largest energy difference, i.e., when $ n_2 \to \infty $.

Hence,

$ \frac{1}{\lambda_{min}} = R_H \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{4} \right) $

$ \lambda_{min} = \frac{4}{R_H} = \frac{4}{1.097 \times 10^7} = 3.64 \times 10^{-7}\ \text{m} $

$ \lambda_{min} = 364\ \text{nm} = 3640\ \text{Å} $

Therefore, the shortest wavelength in the Balmer series is $ 3640\ \text{Å} $.