JEE MAIN - Physics (2023 - 10th April Evening Shift - No. 15)
The variation of stopping potential $$\left(\mathrm{V}_{0}\right)$$ as a function of the frequency $$(v)$$ of the incident light for a metal is shown in figure. The work function of the surface is
_10th_April_Evening_Shift_en_15_1.png)
1.36 eV
18.6 eV
2.98 eV
2.07 eV
Explanation
$$
\begin{aligned}
& e V_0=\frac{h c}{\lambda}-\phi \\\\
& V_0=\frac{h c}{e \lambda}-\left(\frac{\phi}{e}\right)=\frac{h v}{e}-\left(\frac{\phi}{e}\right)
\end{aligned}
$$
When $V_0=0$,
$$ \begin{aligned} \frac{\phi}{e} & =\left(\frac{h v}{e}\right) \\\\ \phi & =h v=\frac{6.626 \times 10^{-34} \times 5 \times 10^{14}}{1.6 \times 10^{-19}} \\\\ & =\frac{6.626 \times 5}{1.6} \times 10^{-1} \\\\ & =\frac{6.626 \times 5}{16} \approx 2.07 \mathrm{eV} \end{aligned} $$
When $V_0=0$,
$$ \begin{aligned} \frac{\phi}{e} & =\left(\frac{h v}{e}\right) \\\\ \phi & =h v=\frac{6.626 \times 10^{-34} \times 5 \times 10^{14}}{1.6 \times 10^{-19}} \\\\ & =\frac{6.626 \times 5}{1.6} \times 10^{-1} \\\\ & =\frac{6.626 \times 5}{16} \approx 2.07 \mathrm{eV} \end{aligned} $$
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