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.

Options

A. 1 V
B. 2 V
C. 3 V
D. 5 V

Answer

2

Explanation

According to Eistein.s photoelectric equation
`V_(0) = (h)/( e) (v - v_(0))`
Substuting the given values, we get
`v_(0) = (6.63 xx 10^(-34)(8.2 xx 10^(14)-3.3 xx 10^(14)))/(1.6 xx 10^(-19)`
2V