The threshold frequency for a certain metal

Threshold frequency of the metal, $v_{0}=3.3 \times 10^{14} \mathrm{~Hz}$

Frequency of light incident on the metal, $v=8.2 \times 10^{14} \mathrm{~Hz}$

Charge on an electron, $e=1.6 \times 10^{-19} \mathrm{C}$

Planck's constant, $h=6.626 \times 10^{-34} \mathrm{Js}$

Cut-off voltage for the photoelectric emission from the metal $=V_{0}$

The equation for the cut-off energy is given as:

$e V_{0}=h\left(v-v_{0}\right)$

$V_{0}=\frac{h\left(v-v_{0}\right)}{e}$

$=\frac{6.626 \times 10^{-34} \times\left(8.2 \times 10^{14}-3.3 \times 10^{14}\right)}{1.6 \times 10^{-19}}=2.0292 \mathrm{~V}$

Therefore, the cut-off voltage for the photoelectric emission is $2.0292 \mathrm{~V}$.