When an electron de-excites from higher orbit in H-atom, two radiations are emitted out in Paschen a — Atomic Structure Chemistry Question

💡 Solution & Explanation

The electron emits two radiations, one in Paschen (terminating at $n=3$) and one in Lyman (terminating at $n=1$). This means it transitioned $n_i \to 3 \to 1$. The radiation in the Lyman series is strictly the $3 \to 1$ transition. $\frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{3^2}\right) = \frac{8R}{9}$. Wavelength $\lambda = \frac{9}{8R}$. Therefore, correct answer is D.