1.

The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is themaximum kinetic energy of photoelectrons emitted?

Answer»

SOLUTION :`V_(0)=1.5V`
`K_("max")=(1)/(2)mv_("max")^(2)=eV_(0)=1.6xx10^(-19)xx1.5=2.4xx10^(-19)J`


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