$\begin {array} {1 1} (a)\;20.292V & \quad (b)\;20292 \times 10^{-4}kV \\ (c)\;202.92V & \quad (d)\;2.0292V \end {array}$

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Ans : (d) 2.0292V

$v_{\circ} = 3.3 \times 10^{14}Hz\: \: \: \: \:v=8.2 \times 10^{14}Hz$

cut off voltage $= V_{\circ}$

So, $eV_{\circ} = h(v-v_{\circ})$

$ \Rightarrow V_{\circ} = h(v – v_{\circ})/e$

$= 6.626 \times 10^{-14} \times (8.2 \times 10^{14} – 3.3 \times 10^{14} ) / 1.6 \times 10^{19}$

$= 2.0292 V$

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