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= 6.6×10−34Js)
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 C1.05×106m/s Energy of incident radiation =hν=hcλ =6.6×10−34Js×3×108m/s200×10−9m=9.90×10−19J….(1)
Given value of work function = 3 eV
Converting 3 eV to Joules =3×1.6×10−19J=4.80×10−19J
Now we know from photoelectric equation,
Incident radiation energy = work function + kinetic energy hν=hν0+12meV2max
By substituting values, 9.90×10−19J=4.80×10−19J+12meV2max12meV2max=5.10×10−1912×9.1×10−31kgV2max=5.10×10−19Vmax=1.05×106m/s