NEET Physics Photoelectric Effect MCQ Question
If the stopping potential for a particular metal is 2 V when exposed to light of frequency 6 x 10^14 Hz, what is the maximum kinetic energy of the photoelectrons emitted?
2 x 10^-19 J
6 x 10^-19 J
4 x 10^-19 J
8 x 10^-19 J
Correct Answer
Detailed Explanation
The maximum kinetic energy K_max of the photoelectrons is given by eV_0, where e is the charge of the electron (1.6 x 10^-19 C) and V_0 is the stopping potential. Thus, K_max = 1.6 x 10^-19 C x 2 V = 3.2 x 10^-19 J.
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