The threshold frequency of 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 photoelectric emission. Given Planck's constant, `h=6.62xx10^(-34)Js`.

Options

A. 2 V
B. 4 V
C. 6 V
D. 8 V

Answer

A

Explanation

(a) : Given threshold freqency, `v_(0)=3.3xx10^(14)Hz`
Frequency of incident light, `v=8.2xx10^(14)Hz`
As `eV_(0)=h(v-v_(0)) or V_(0)=(h(-v_(0)))/(e)`
`V_(0)=(6.63xx10^(-34)(8.2xx10^(14)-3.3xx10^(14)))/(1.6xx10^(-19))=2V`