A photon collides with a stationary hydrogen atom in ground state in-elastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom in-elastically with an energy of 15 eV. What will be observed by the detector?

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2 photons of energy 10.2 eV

2 photons of energy 1.4 eV

One photon of energy 10.2 eV and an electron of energy 1.4 eV

One photon of energy 10.2 eV and another photon of energy 1.4 eV

answer is C.

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Detailed Solution

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

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