Question

The maximum kinetic energy of the photoelectrons is found to be `6.63xx10^(-19)J`, when the metal is irradiated with a radiation of frequency `2xx10^(15)`Hz. The threshold frequency of the metal is about:

Answer 1

Correct Answer - `1xx10^(-15)s^(-1)`
Kinetic energy (KE) `=h(v-v_(0))`
`v-v_(0)=(KE)/(h) or v_(0)=v-(KE)/(h)`
`v_(0)=(2xx10^(15)s^(-1))-((6.63xx10^(-19)J))/((6.626xxx10^(-34)Js))`
`=(2xx10^(15)s^(-1))-(1xx10^(15)s^(-1))=1xx10^(15)s^(-1)`