The threshold frequency for a certain metal is `3.3xx10^(14) Hz.` If light of frequency `8.2xx10^(1)` Hz (hertz) is incident on the metal , find the cut - off voltage for photoelectric emission . Given Planck's constant `h = 6.625xx10^(-34)` Js (joule second ) . Charge of electron `e = 1.6 xx10^(-19)` C (coulomb).

Options

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

Answer

B

Explanation

According to Einstein.s photoelectric equation
`eV_s = hupsilon - hupsilon_0`
where, `upsilon ` = Incident frequency
`V_s` = Cut - off or stopping potential
or `V_s = (h)/(e) (upsilon-upsilon_0)`
Substituting the given values, we get
`V_s=(6.63 xx10^(-34)(8.2 xx10^(14)-3.3 xx10^(14)))/(1.6 xx10^(-19))~~2V`