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

Answer»

SOLUTION :As `K_(max)=eV_(0) and` cut-off voltage `V_(0)=1.5V`
`THEREFORE K_(max)=1.5xx1.6xx10^(-19)J=2.4xx10^(-19)J or 1.5eV`.


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