1. The stopping potential in an experiment on photoelectric effect is 2 v.What is the maximum kinetic energy of the photoelectrons emitted?
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KE max = 2ev ,
which is 2* 1.6 * 10^-19 joules..
which is 3.2 * 10^-19joules
mark as brainliest please
which is 2* 1.6 * 10^-19 joules..
which is 3.2 * 10^-19joules
mark as brainliest please
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