The threshold frequency for certain metal is $3.3\times {10}^{14}Hz$. If light of $8.2\times {10}^{14}Hz$ frequency is incident on the metal, then the cut-off voltage of the photoelectric current will be :

In photoelectric effect, energy is conserved.

In photoelectric effect, Einstein's equation is given by

Photon energy $=$ kinetic energy of electron $+$ work function

i.e. $hv=e{V}_s+h{v}_0$

where, $V_s$ is the stopping potential and $v_0$ is the threshold frequency.

$\therefore V_s=\frac{h}{e}\left(v-v_0\right)$              .....(i)

 Given, $v_0=3.3\times {10}^{14}Hz$, $v=8.2\times {10}^{14}Hz$

            $h=6.6\times {10}^{-34}J-s$, $e=1.6\times {10}^{-19}C$

Substituting these values in equation (i), we get

$V_s=\frac{6.6\times {10}^{-34}\left(8.2\times {10}^{14}-3.3\times {10}^{14}\right)}{1.6\times {10}^{-19}}$

$V_s=\frac{6.6\times 4.9}{1.6}\times {10}^{-1}=2V$

$\Rightarrow V_s=2V$