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. `3 V`
C. `5 V`
D. `1 V`

Answer

A

Explanation

`V_(0) = (E - V)/(e ) = (h (v - v_(0)))/( e )`
` = (6.62 xx 10^(-34) ( 8.2 xx 10^(14) - 3.3 xx 10^(14)))/(1.6 xx 10^(-19))`
` = (6.62 xx 4.9 xx 10^(-1))/(1.6)`
`V_(0) = 2 volt`