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Question

Photoelectrons are ejected from the surface of a metal having work function 4.5 eV. Then the impulse transmitted to the surface of the metal when the electron flies off due to collision of light quanta of energy 4.9eV.

A
3.93×1025Kgms1
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B
1.47×1025Kgms1
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C
3.43×1025Kgms1
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D
4.62×1025Kgms1
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Solution

The correct option is C 3.43×1025Kgms1
Work function of metal =4.5eV
Energy of photons of light =4.9eV
Now,
Energy imparted to the electrons =(4.94.5)eV
=0.4eV
Now to find the velocity,

12mv2=0.4eV

9.1×1031×v2=0.8×1.6×1019
v=3.750×105 m/s
So,
Impulse transmitted to surface of metal when an electron flies off = mv
=9.1×1031×3.750×105kgms1
3.412×1025kgms1
3.43×1025kgms1
So, the answer is option (C).

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