Question on: JAMB Physics - 2000
Light of energy 5eV falls on a metal of work function 3eV and electrons are liberated. The stopping potential is
A
1.7V
B
2.0V
C
8.0V
D
15.0V
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Correct Option: B
For a photoemitted electron,
eV = K.E.max, where V = stopping potential and K.E. = maximum kinetic energy of the photo emitted electron given by K.E. = Energy of incident rad - Work - function
= 5ev - 3ev = 2ev
∴ eV = 2eV
V = 2eV/e = 2V
∴ Stopping Potential = 2.0V
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