The threshold frequency for a certain metal is `3.3xx10^(14)` Hz.If light of frequency `8.2xx10^(14)`Hz is incident on the metal ,predict the cutoff voltage for the photoelectric emission.

Explanation

Here `V_(0)=3.3xx10^(14)Hz,h=6.63xx10^(-34)Js`
`v=8.2xx10^(14)Hz`
`V_(0)=?`
`implies` Maximum kinetic energy of photon electron,
`(1)/(2)mv_(max)^(2)=hv-phi_(0)`
`therefore eV_(0)=hv-hv_(0)`
`therefore V_(0)=(h)/(e )(v-v_(0))`
`=(6.63xx10^(-34)xx(8.2xx10^(14)-3.3xx10^(14)))/(1.xx10^(-19))`
`=20.30xx10^(-1)`
`therefore V_(0)=2.03 V`