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

The wavelength of a photon emitted during a transition in the hydrogen atom is given by the Rydberg formula: $$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$, where $$R_H = 1.097 \times 10^7$$ m$$^{-1}$$ is the Rydberg constant, $$n_1$$ is the lower energy level, and $$n_2$$ is the higher energy level.

For the transition from $$n = 2$$ to $$n = 1$$ (Lyman series): $$\frac{1}{\lambda} = R_H\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R_H\left(1 - \frac{1}{4}\right) = \frac{3R_H}{4}$$.

$$\frac{1}{\lambda} = \frac{3 \times 1.097 \times 10^7}{4} = \frac{3.291 \times 10^7}{4} = 8.228 \times 10^6 \; \text{m}^{-1}$$

$$\lambda = \frac{1}{8.228 \times 10^6} = 1.215 \times 10^{-7}$$ m $$= 121.5$$ nm.

This is approximately 121.8 nm, which falls in the ultraviolet region and is the first line of the Lyman series.