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Q.

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∴ Kmax⁡hυ=25×100%=40%
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