A photon collides with a stationary hydrogen atom in the ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of microsecond another photon collides with same hydrogen atom inelastically with an energy

Question

A photon collides with a stationary hydrogen atom in the ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of microsecond another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector? [2005]

Smart Achievers · Accepted Answer

(1)

Initially, a photon of energy 10.2 eV collides inelastically with a hydrogen atom in the ground state. For hydrogen atom,

$E_{1} = - 13.6 eV, E_{2} = - \frac{13.6}{4} eV = - 3.4 eV$

$∴$ $E_{2} - E_{1} = 10.2 eV$

The electron of the hydrogen atom will jump to the second orbit after absorbing the photon of energy 10.2 eV. Another photon of energy 15 eV strikes the hydrogen atom inelastically. This energy is sufficient to knock out the electron from the atom as the ionisation energy is 13.6 eV. The remaining energy, $15 - 13.6 = 1.4 eV$ is left, which is released by the second photon.

Solution

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

2 2 photon of energy of 1.4 eV

3 2 photon of energy 10.2 eV

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

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