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 cut-off voltage for the photoelectric emission.

Explanation

Given, `v_(0)=3.3xx10^(14)Hz , v =8.2xx10^(14)Hz , e =1.6xx10^(-19)c, KE = eV_(0)= hv - hv_(0)`
`V_(0)=(h(v-v_(0)))/(e )=(6.63xx10^(-34)xx(8.2-3.3)xx10^(14))/(1.6xx10^(-19))=(6.63xx10^(-34)xx10^(14)xx4.9)/(1.6xx10^(-19)) therefore V_(0) = 2.03 V`.