The graph of stopping potential $\left(\mathrm{V}_{s}\right)$ against frequency $(v)$ of incident radiation…

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
  1. $\phi_{\mathrm{P}}=\phi_{\mathrm{Q}}$
  2. $v_{0}^{\prime} < v_{0}$
  3. $\phi_{\mathrm{P}} < \phi_{\mathrm{Q}}$
  4. $\phi_{\mathrm{P}}>\phi_{\mathrm{Q}}$

Solution

$\phi=\mathrm{hv}_{\mathrm{o}} \quad \therefore \phi \propto \mathrm{v}_{\mathrm{o}}$ $v_{\mathrm{o}} < v_{\mathrm{o}}^{\prime} \quad \therefore \phi_{\mathrm{p}} < \phi_{\mathrm{Q}}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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