In a photoelectric experiment work function of the metal surface is 3 eV and stopping potential is 2 volt then what percent of the energy of the photon of incident light appears as kinetic energy of fastest moving photo-electron?
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a
60%
b
40%
c
30%
d
20%
answer is B.
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Detailed Solution
Kmax=2 eV and ϕ=3 eV∴ Energy of incident photon, hυ=ϕ+Kmax=5 eV∴ Kmaxhυ=25×100%=40%