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Question

Work function of a metal is 3eV. An electromagnetic radiation of 200 nm is incident on metal surface, calculate the maximum speed of photoelectrons ejected. (Take the value of h= 6.6×1034Js)

A
2.05×106m/s
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B
1.75×106m/s
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C
1.05×106m/s
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D
2.20×106m/s
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Solution

The correct option is C 1.05×106m/s
Energy of incident radiation =hν=hcλ
=6.6×1034J s×3×108m/s200×109m=9.90×1019J.(1)
Given value of work function = 3 eV
Converting 3 eV to Joules =3×1.6×1019J=4.80×1019J
Now we know from photoelectric equation,
Incident radiation energy = work function + kinetic energy
hν=hν0+12meV2max
By substituting values,
9.90×1019J=4.80×1019J+12meV2max12meV2max=5.10×101912×9.1×1031kg V2max=5.10×1019Vmax=1.05×106m/s

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