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Question

The graph shown below depicts plot of photocurrent versus anode potential for a cathode with 4 eV work function. The energy of the incident photon is
1068243_c78a16c0021249e4975ccd3e8667697f.png

A
6 eV
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B
4 eV
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C
2 eV
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D
8 eV
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Solution

The correct option is A 6 eV

It is given that

Work function

ϕ=4eV

V=2V

From Einstein’s relation

hν=ϕ+Ek

Where hν is the energy of photons. And Ek is the kinetic energy

hν=(4+2)eV

hν=6eV


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