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Question

Light of wavelength 4000A falls on a photosensitive metal and a negative 2V potential stops the emitted electrons. The work function of the material (in eV) is approximately (h=6.6×1034Js,e=1.6×1019 C,c=3×108 ms1)

A
1.1
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B
2
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C
2.2
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D
3.1
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Solution

The correct option is A 1.1
Energy of incident light E(eV)=123754000=3.09 eV
Stopping potential is-2V so Kmax;=2eV
Hence by using E=W0+Kmax;W=1.09 eV1.1 eV

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