The wavelength of photon emitted by a hydrogen atom when an electron makes a transition from $n = 2$ to $n = 1$ state is.

(Take the value of the Rydberg constant, $R = 1.0973 × 10^7\, m^{-1}$)

121.5 nm

The correct answer is Option (1) → 121.5 nm

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

$\frac{1}{\lambda} = 1.0973 \times 10^{7} \left(1 - \frac{1}{4}\right)$

$\frac{1}{\lambda} = 1.0973 \times 10^{7} \cdot \frac{3}{4}$

$\frac{1}{\lambda} = 8.22975 \times 10^{6} \ \text{m}^{-1}$

$\lambda = \frac{1}{8.22975 \times 10^{6}}$

$\lambda = 1.215 \times 10^{-7} \ \text{m}$

$\lambda = 121.5 \ \text{nm}$

The wavelength is $1.215 \times 10^{-7} \ \text{m}$ or $121.5 \ \text{nm}$.