In the atomic hydrogen line spectrum, the wavelength of the first line of the Lyman series is

(Given: Rydberg constant $R = 1.097×10^7\, m^{-1}$)

122 nm

The correct answer is Option (2) → 122 nm

For the Lyman series, the electronic transition ends at $n_1 = 1$. The first line corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.

The wavelength is given by the Rydberg formula:

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

Substitute values:

$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$

$\frac{1}{\lambda} = 1.097 \times 10^7 \left( 1 - \frac{1}{4} \right) = 1.097 \times 10^7 \times \frac{3}{4}$

$\frac{1}{\lambda} = 8.2275 \times 10^6$

$\lambda = \frac{1}{8.2275 \times 10^6} \approx 1.215 \times 10^{-7} \ \text{m} = 121.5 \ \text{nm}$