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Photoelectric Effect

A photon of e...

Question

A photon of energy $70 e V$ collides with $H e^{+}$ ion. Due to this collision, $H e^{+}$ atom gets ionized. The maximum kinetic energy of the emitted electron is

A

13.6 e V

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B

15.6 e V

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C

27.2 e V

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D

31.2 e V

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Solution

The correct option is B $15.6 e V$
From Bohr's theory for a hydrogen like atom
$E_{n} = - \frac{Z^{2}}{n^{2}} 13.6 e V$
For $H e^{+}, Z = 2 a n d n = 1$
$\Rightarrow E_{1} = - 54.4 e V$

By conserving energy

K.E of electron = Energy of Photon - $E_{1}$
$= 70 - 54.4$
$= 15.6 e V$

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