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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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a
– 4.2 V
b
– 10.2 V
c
– 17.8 V
d
+9.4 V

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detailed solution

Correct option is B

E=W0+eV0For hydrogen atom,  E=+13.6 eV ∴ + 13.6 = 3.4 + eV0 ⇒V0=(13.6−3.4) eVe=10.2 VPotential at anode = – 10.2 V
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