1. Tardigrade
  2. Question
  3. Physics
  4. A photon collides with a stationary hydrogen atom in ground state inelastically. The energy of the colliding photon is 10.2 eV . After a time interval of the order of a microsecond, another photon collides with the same hydrogen atom inelastically with an energy of 15 eV . What will be observed by the detector?

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Q. A photon collides with a stationary hydrogen atom in ground state inelastically. The energy of the colliding photon is $10.2 \, eV$ . After a time interval of the order of a microsecond, another photon collides with the same hydrogen atom inelastically with an energy of $15 \, eV$ . What will be observed by the detector?

NTA AbhyasNTA Abhyas 2022

Solution:

The first photon will excite the hydrogen atom (in ground state) to first excited state (as $E_{2}-E_{1}=\text{10.2} \, eV$ ). Hence, during de-excitation a photon of $\text{10.2} \, eV$ will be released.
The second photon of energy $15 \, eV$ can ionise the atom. Hence the balance energy i. e., $\left(15 - \text{13.6}\right) \, eV=\text{1.4} \, eV$ is retained by the electron. Therefore, by the second photon an electron of energy $\text{1.4} \, eV$ will be released.