The smallest wavelength in the spectral lines of Paschen series in the hydrogen atom is

(Given: $R = 1.097 × 10^7 m^{-1}$)

8204 Å

The correct answer is Option (3) → 8204 Å

For hydrogen atom, the wavelength of spectral lines is given by Rydberg formula:

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

For Paschen series, $n_1 = 3$, $n_2 = 4,5,6,...$

The smallest wavelength corresponds to the largest energy difference → $n_2 \to \infty$

$\frac{1}{\lambda_\text{min}} = R \left( \frac{1}{3^2} - 0 \right) = \frac{R}{9}$

$\lambda_\text{min} = \frac{9}{R}$

Given $R = 1.097 \times 10^7 \, \text{m}^{-1}$

$\lambda_\text{min} = \frac{9}{1.097 \times 10^7} \approx 8.204 \times 10^{-7} \, \text{m}$