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Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 3.4 eV. In order to reduce the photo-current to zero the voltage of the anode relative to the cathode must be made

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– 4.2 V

– 10.2 V

– 17.8 V

+9.4 V

answer is B.

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Detailed Solution

$E = W_{0} + {eV}_{0}$

For hydrogen atom, $E = + 13 .6 eV$
$∴ + 13.6 = 3.4 + {eV}_{0}$

$\Rightarrow V_{0} = \frac{(13 .6 - 3 .4) eV}{e} = 10 .2 V$

Potential at anode = – 10.2 V

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