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

Explanation

`upsilon_(0)=3.3xx10^(14)Hz, upsilon=8.2xx10^(14)Hz`
`(1)/(2)mv_("max")^(2)=h(upsilon-upsilon_(0))`, But `eV_(0)=(1)/(2)mv_("max")^(2)`
`eV_(0)=h(upsilon-upsilon_(0))=6.6xx10^(-34)[8.2xx10^(14)-3.3xx10^(14)]`
`V_(0)=(6.6xx10^(34)xx4.9xx10^(14))/(1.6xx10^(-19))=20.2xx10^(-1)V=2V`