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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? |
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Answer» (K.E)max = q x Vo ;where, Vo = stopping potential = e x 4V ; e = charge on an electron (K.E)max = 4eV The answer is: 4eV |
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