Photons of same energy were allowed to strike on two different samples of hydrogen atoms, one having — Atomic Structure Chemistry Question

💡 Solution & Explanation

Let the incident photon energy be $E$. The kinetic energy of photoelectrons from the ground state is $K_1 = E - 13.6$. The kinetic energy from the $n^{th}$ excited state is $K_2 = E - \frac{13.6}{n^2}$. The difference is $K_2 - K_1 = 13.6 - \frac{13.6}{n^2} = 12.75$ eV. This yields $\frac{13.6}{n^2} = 0.85 \implies n^2 = \frac{13.6}{0.85} = 16 \implies n = 4$. Therefore, correct answer is 4.