Threshold frequency of the metal, \(\nu_{0}\) = 3.3 x ${10}^{14}$ Hz
Frequency of light incident on the metal, \(\nu\) = 8.2 x ${10}^{14}$ Hz
Charge on an electron, e = 1.6 x ${10}^{-19}$ C
Planck's constant, h = 6.626 x ${10}^{-34}$ Js
Cut-off voltage for the photoelectric emission from the metal = \(V_{0}\)
The equation for the cut-off energy is given as:
$\begin{array}{l}e{V}_0=h(v-v_0)\\ V_0=\frac{h(v-v_0)}{e}\\ =\frac{6.626\times {10}^{-34}\times (8.2\times {10}^{14}-3.3\times {10}^{14})}{1.6\times {10}^{-19}}=2.0292\mathrm{\hspace{0.17em}\; V.}\end{array}$