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. 4.9 V
B. 3.0 v
C. 2.0 V
D. 1.0 V

Answer

C

Explanation

`K_("max") = h(v-v_0)`
`implies eV_0 = 6.6 xx10^(-34) (8.2 - 3.3 ) xx10^(14)`
`implies v_0 = (6.6 xx10^(-34) xx4.9xx 10^(14))/(1.6 xx10^(-19))=2.021 V~~ 2V`