The cut-off voltage in a photoelectric experiment is 3V. Then the maximum KE of photo-electrons emitted is
A
3 V
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B
3 eV
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C
6 eV
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D
9 eV
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Solution
The correct option is B 3 eV Cut off voltage is the minimum voltage applied across the plates that even the electrons ejected with minimum kinetic energy could not reach the other plate. So, From the definition, Cut off voltage =3V Work done on the charge = Kinetic energy of the photons ⇒3V×1e=K.E. ⇒K.E.=3eV