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the maximum kinetic energy of photoelectron is 2.8eV , what is the value of its stopping potential

The Stopping Potential of an electron can be calculted from its maximum K.E by the following expression

${V}_{0}=\frac{K.E}{e}wheree=chargeonanelectron\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Here,\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{V}_{0}=\frac{2.8eV}{e}=2.8V$

Regards.

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