1.

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

Answer» Give, `V_(0)=1.5 V, e = 1.6xx10^(-19)C, KE_("max")=eV_(0)=1.6xx10^(-19)xx1.5=2.4xx10^(-19)J`.


Discussion

No Comment Found

Related InterviewSolutions