1.

In a photoelectric experiment, the potential required to stop the ejection of electrons from cathode is 4V. What is the value of maximum kinetic energy of emitted Photoelectrons?

Answer»

(K.E)max = q x V ;where, Vo = stopping potential

= e x 4V ; e = charge on an electron

(K.E)max = 4eV

The answer is: 4eV



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