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In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15} \mathrm{~V}$ s. Calculate the value of Planck's constant.
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For a graph of $\mathrm{V}_0$ vs $v$, slope gives $\frac{\mathrm{h}}{\mathrm{e}}$
$$
\begin{aligned}
\therefore & \frac{\mathrm{h}}{\mathrm{e}}=4.12 \times 10^{-15} \mathrm{~V} / \mathrm{s} . \\
\Rightarrow \quad \mathrm{h} &=\mathrm{e} \times \mathrm{slope}=1.6 \times 10^{-19} \times 4.12 \times 10^{-15} \\
&=6.59 \times 10^{-34} \mathrm{Js}
\end{aligned}
$$
$$
\begin{aligned}
\therefore & \frac{\mathrm{h}}{\mathrm{e}}=4.12 \times 10^{-15} \mathrm{~V} / \mathrm{s} . \\
\Rightarrow \quad \mathrm{h} &=\mathrm{e} \times \mathrm{slope}=1.6 \times 10^{-19} \times 4.12 \times 10^{-15} \\
&=6.59 \times 10^{-34} \mathrm{Js}
\end{aligned}
$$
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