Photons of same energy were allowed to strike on two different samples of hydrogen atoms, one having — Atomic Structure Chemistry Question
Question
Photons of same energy were allowed to strike on two different samples of hydrogen atoms, one having each atom in ground state and other, in a particular excited state of orbit number $n$. The photonic beams ionize the hydrogen atoms in both the samples. If the difference in maximum kinetic energy of emitted photoelectron from both the samples is 12.75 eV, the value of $n$ is
💡 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.