Structure of Atom Part 2

Concept:The atomic spectrum of hydrogen arises from electron transitions between energy levels, with frequency given by the Rydberg formula for hydrogen-like species.Explanation:Statement I is true. When electric discharge is passed through gaseous hydrogen, $\mathrm{H}_{2}$ molecules dissociate into atoms. These atoms get excited and, upon returning to lower energy levels, emit radiation of discrete frequencies, producing a line spectrum. Statement II is also true. Use the Rydberg formula: $\tilde{\nu} = R Z^{2}\left(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right)$. For the second line of the Balmer series of $\mathrm{He}^{+}$: Balmer series has $n_{1}=2$, second line means $n_{2}=4$. With $Z=2$, $\tilde{\nu}_{\mathrm{He}^{+}} = R(2)^{2}\left(\frac{1}{2^{2}} - \frac{1}{4^{2}}\right) = 4R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{4}$. For the first line of the Lyman series of hydrogen: Lyman series has $n_{1}=1$, first line means $n_{2}=2$. With $Z=1$, $\tilde{\nu}_{\mathrm{H}} = R(1)^{2}\left(\frac{1}{1^{2}} - \frac{1}{2^{2}}\right) = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$. Thus, $\tilde{\nu}_{\mathrm{He}^{+}} = \tilde{\nu}_{\mathrm{H}}$, so their frequencies are the same.Answer:Both Statement I and Statement II are true. Correct option: A.