Physics, asked by zaidkhan787, 2 months ago

the maximum K.E. of a photo electron is 3ev. what is its stopping potential

Answers

Answered by shilpashewkani

6

Answer:

Solution :

Since, Kmax=eV0

3×1.6×10-19=1.6×10-9V0[∵1eV=1.6×10-19J]

∴ Stopping potential V0=3V.

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