MHT CET · Physics · Dual Nature of Matter
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
- A \(\phi_{\mathrm{P}}=\phi_{\mathrm{Q}}\)
- B \(v_{0}^{\prime} < v_{0}\)
- C \(\phi_{\mathrm{P}} < \phi_{\mathrm{Q}}\)
- D \(\phi_{\mathrm{P}}>\phi_{\mathrm{Q}}\)
Answer & Solution
Correct Answer
(C) \(\phi_{\mathrm{P}} < \phi_{\mathrm{Q}}\)
Step-by-step Solution
Detailed explanation
(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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