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The ratio of slopes of $K_{\max }$ vs $V$ and $V_{0}$ vs $v$ curves in the photoelectric effects gives $\left(v=\right.$ frequency,$K_{\max }=$ maximum kinetic energy, $v_{0}=$ stopping potential $)$
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charge of electron
$\begin{array}{l}
\mathrm{hv}=\mathrm{hv}_{0}+\mathrm{ev}_{0} \\
\mathrm{v}_{0}=\frac{\mathrm{h}}{\mathrm{e}} \mathrm{v}-\frac{\mathrm{h}}{\mathrm{e}} \mathrm{v}_{0}
\end{array}$
On comparing this equation with the straight line equation, i.e $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ The slope of $\mathrm{v}_{0}$ vs $\mathrm{v}$ is ( $v_{0}$ is stopping potential)
$(\text { slope })_{1}=\frac{\mathrm{h}}{\mathrm{e}}$
Likewise,
$\mathrm{hv}=\mathrm{hv}_{0}+\mathrm{K}_{\max }$
$\text { or } \quad \mathrm{K}_{\max }=\mathrm{h} \mathrm{v}-\mathrm{h} \mathrm{v}_{0}$
Thus, slope of $K_{\max }$ vs $v$ is
$(\text { slope })_{2}=\mathrm{h} \therefore \frac{(\text { slope })_{2}}{(\text { slope })_{1}}=\frac{\mathrm{h}}{\mathrm{h} / \mathrm{e}}=\mathrm{e}$
\mathrm{hv}=\mathrm{hv}_{0}+\mathrm{ev}_{0} \\
\mathrm{v}_{0}=\frac{\mathrm{h}}{\mathrm{e}} \mathrm{v}-\frac{\mathrm{h}}{\mathrm{e}} \mathrm{v}_{0}
\end{array}$
On comparing this equation with the straight line equation, i.e $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ The slope of $\mathrm{v}_{0}$ vs $\mathrm{v}$ is ( $v_{0}$ is stopping potential)
$(\text { slope })_{1}=\frac{\mathrm{h}}{\mathrm{e}}$
Likewise,
$\mathrm{hv}=\mathrm{hv}_{0}+\mathrm{K}_{\max }$
$\text { or } \quad \mathrm{K}_{\max }=\mathrm{h} \mathrm{v}-\mathrm{h} \mathrm{v}_{0}$
Thus, slope of $K_{\max }$ vs $v$ is
$(\text { slope })_{2}=\mathrm{h} \therefore \frac{(\text { slope })_{2}}{(\text { slope })_{1}}=\frac{\mathrm{h}}{\mathrm{h} / \mathrm{e}}=\mathrm{e}$
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