An electron revolving round H-nucleus in ground state absorbs 10.2 eV energy. Its angular momentum i — Atomic Structure Chemistry Question

💡 Solution & Explanation

The ground state is $n=1$ (-13.6 eV). Absorbing 10.2 eV excites it to $n=2$ (-3.4 eV). The change in angular momentum is $\Delta L = \frac{2h}{2\pi} - \frac{1h}{2\pi} = \frac{h}{2\pi}$. Therefore, correct answer is A.