1. Tardigrade
  2. Question
  3. Physics
  4. A photon collides with a stationary hydrogen atom in ground state inelastieally. 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 inelastieally 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 inelastieally. 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 inelastieally with an energy of 15 eV. What will be observed by the detector ?

IIT JEEIIT JEE 2005

Solution:

The first photon will excite the hydrogen atom (in ground state) to first excited state (as $E_2-E_1 = 10.2eV).$ Hence, during de-excitation a photon of 10.2 eV will be released. The second photon of energy 15 eV can ionise 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.
$\therefore $ Correct answer is (c).