A hydrogen atom emits ultraviolet radiation having a wavelength of...

The wavelength of the emitted photon is given by the Rydberg formula:

\dfrac{1}{\lambda} = R_H \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)

Since the emitted radiation is in the ultraviolet region, it belongs to the Lyman series, which corresponds to transitions to the ground state ($n_1 = 1$).

Substituting $\lambda = 102.5 \text{ nm} = 102.5 \times 10^{-9} \text{ m} and$R_H = 1.097 \times 10^7 \text{ m}^{-1}$:<br><br>\dfrac{1}{102.5 \times 10^{-9}} =$1.097 \times 10^7 \left( 1 - \dfrac{1}{n_2^2} \right)$<br><br>$9.756 \times 10^6 = 1.097 \times 10^7 \left( 1 - \dfrac{1}{n_2^2} \right)<br><br>1 - $\dfrac{1}{n_2^2} = \dfrac{9.756 \times 10^6}{1.097 \times 10^7} \approx$0.889$<br><br>\dfrac{1}{n_2^2} =$1 - 0.889 = 0.111 \approx $\dfrac{1}{9}<br><br>$n_2^2 = 9 \Rightarrow n_2 = 3$<br><br>Thus, the quantum numbers of the states involved are$1$and$3$.$