Physics - Modern Physics Question with Solution | TestHub

The first photon will excite the hydrogen atom (in ground state) to first excited state (as ${\mathrm{E}}_2-{\mathrm{E}}_1=\text{10.2} \mathrm{eV}$). Hence, during de-excitation a photon of $\text{10.2} \mathrm{eV}$ will be released.

The second photon of energy $15 \mathrm{eV}$ can ionise the atom. Hence the balance energy i. e., $\left(15-\text{13.6}\right) \mathrm{eV}=\text{1.4} \mathrm{eV}$ is retained by the electron. Therefore, by the second photon an electron of energy $\text{1.4} \mathrm{eV}$ will be released.