Modern Physics 1 Question 19

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

(a) 2 photons of energy 10.2eV

(2005,2M)

(b) 2 photons of energy 1.4eV

(c) One photon of energy 10.2eV and an electron of energy 1.4eV

(d) One photon of energy 10.2eV and another photon of energy 1.4eV

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Solution:

  1. The first photon will excite the hydrogen atom (in ground state) to first excited state (as E2E1=10.2eV ). Hence, during de-excitation a photon of 10.2eV will be released. The second photon of energy 15eV can ionise the atom. Hence, the balance energy i.e. (1513.6)eV=1.4eV is retained by the electron. Therefore, by the second photon an electron of energy 1.4eV will be released.


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