What is the frequency of the second line of the Paschen series in the spectrum of ion? — Atomic Structure Chemistry Question

💡 Solution & Explanation

The Paschen series ends at $n_1=3$. The second line corresponds to a transition from $n_2=5$ to $n_1=3$. For $He^+$, $Z=2$. $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{3^2} - \frac{1}{5^2}\right) = R(4)\left(\frac{1}{9} - \frac{1}{25}\right) = 4R\left(\frac{16}{225}\right) = \frac{64 R}{225}$. The frequency $\nu = \frac{C}{\lambda} = C \times \frac{1}{\lambda} = \frac{64 R C}{225}$. Therefore, correct answer is A.