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The graph of stopping potential $\left(\mathrm{V}_{s}\right)$ against frequency $(v)$ of incident radiation is
plotted for two different metals 'P' and 'Q' as shown in the graph. $\phi_{\mathrm{p}}$ and $\phi_{\mathrm{q}}$ are work functions of $\mathrm{P}$ and $\mathrm{Q}$ respectively, then
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plotted for two different metals 'P' and 'Q' as shown in the graph. $\phi_{\mathrm{p}}$ and $\phi_{\mathrm{q}}$ are work functions of $\mathrm{P}$ and $\mathrm{Q}$ respectively, then
Solution:
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Verified Answer
The correct answer is:
$\phi_{\mathrm{P}} < \phi_{\mathrm{Q}}$
(B)
$\phi=\mathrm{hv}_{\mathrm{o}} \quad \therefore \phi \propto \mathrm{v}_{\mathrm{o}}$
$v_{\mathrm{o}} < v_{\mathrm{o}}^{\prime} \quad \therefore \phi_{\mathrm{p}} < \phi_{\mathrm{Q}}$
$\phi=\mathrm{hv}_{\mathrm{o}} \quad \therefore \phi \propto \mathrm{v}_{\mathrm{o}}$
$v_{\mathrm{o}} < v_{\mathrm{o}}^{\prime} \quad \therefore \phi_{\mathrm{p}} < \phi_{\mathrm{Q}}$
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