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Question

The electric field of light wave is given as E=103 cos(2π x5×1072π×6×1014 t)^xNC
This light falls on a metal plate of work function 2 eV. The stopping potential of the photo-electrons is

Given, E(in eV)=12375λ(in Å)

A
2.48 V
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B
0.72 V
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C
0.48 V
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D
2.0 V
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Solution

The correct option is C 0.48 V
Given:
ω=2π×6×1014
f=6×1014 Hz

Wavelength
λ=cf=3×1086×1014=0.5×106 m=5000 Å

E=12375λ (in Å)

Now, E=123755000=2.48 eV

Given that W=2eV

Using E=W+eVs

2.48 eV=2 eV+eVs

Vs=0.48 V

Hence option (C) is correct.

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