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In a photoelectric experiment, a graph is drawn with stopping potential along $\mathrm{Y}$-axis and the frequency of the incident light along $\mathrm{X}$-axis. If the graph is a straight line which makes an angle $\theta$ with Y-axis, then $\tan \theta=$ (h-planck's constant, e-charge of electron)
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The correct answer is:
$\frac{\mathrm{h}}{\mathrm{e}}$
From the photoelectric equation
$\begin{aligned} & \mathrm{eV}_{\mathrm{s}}=\mathrm{hv}-\phi \\ & \mathrm{V}_{\mathrm{s}}=\frac{\mathrm{hv}}{\mathrm{e}}-\frac{\phi}{\mathrm{e}}\end{aligned}$
By compairing above equation with $\mathrm{y}=\mathrm{mx}+\mathrm{c}$
slope $=\mathrm{m}=\tan \theta=\frac{\mathrm{h}}{\mathrm{e}}$
$\begin{aligned} & \mathrm{eV}_{\mathrm{s}}=\mathrm{hv}-\phi \\ & \mathrm{V}_{\mathrm{s}}=\frac{\mathrm{hv}}{\mathrm{e}}-\frac{\phi}{\mathrm{e}}\end{aligned}$
By compairing above equation with $\mathrm{y}=\mathrm{mx}+\mathrm{c}$
slope $=\mathrm{m}=\tan \theta=\frac{\mathrm{h}}{\mathrm{e}}$
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