1.

The photoelectric cutoff voltage in a certain expertiment is 1.5 V.What is the maximum kinetic energy of photoelectrons emitted?

Answer»

Solution :Here `V_(0)=15V,e=1.6xx10^(-19)C`
`K_(max)=eV_(0)=1.6xx10^(-19)xx1.5=2.4xx10^(-19)J`


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