Question

A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition?

Answer 1

Wavelength of ultraviolet radiation, $\lambda$ = 102.5 nm = 102,5 $\times$10^$-9$ m
Rydberg's constant, R = 1.097$\times$ 10⁷ m^$-1$
Since the emitted light lies in ultraviolet range, the lines will lie in lyman series.
Lyman series is obtained when an electron jumps to the ground state (n_​1 = 1) from any excited state (n_​2).
Wavelength of light $\left(\lambda \right)$ is given by
$$\begin{gathered} \frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\ \mathrm{Here},R=\mathrm{Rydberg}\mathrm{constant} \\ \Rightarrow \frac{1}{102.5\times {10}^{-9}}=1.097\times {10}^7\left(\frac{1}{1^2}-\frac{1}{n_2^2}\right) \\ \Rightarrow \frac{{10}^9}{102.5}=1.097\times {10}^7\left(1-\frac{1}{n_2^2}\right) \\ \Rightarrow \frac{{10}^2}{102.5}=1.097\left(1-\frac{1}{n_2^2}\right) \\ \Rightarrow 1-\frac{1}{n_2^2}=\frac{100}{102.5\times 1.097} \\ \Rightarrow \frac{1}{n_2^2}=1-\frac{100}{102.5\times 1.097} \\ \Rightarrow n_2=\sqrt{9.041}=3 \end{gathered}$$
The transition will be from 1 to 3.