JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2024
- Hard
- MCQ
When a metal surface is illuminated by light of wavelength \(\lambda\), the stopping potential is \(8 \mathrm{~V}\). When the same surface is illuminated by light of wavelength \(3 \lambda\), stopping potential is \(2 \mathrm{~V}\). The threshold wavelength for this surface is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(9 \lambda\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\phi+\mathrm{K}_{\max }\)
\(\phi=\frac{\mathrm{hc}}{\lambda_0}\)
\(\mathrm{K}_{\max }=\mathrm{eV}_0\)
\(8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . \text { (i) }\)
\(2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . . \text { (ii) }\)
\(\text {on solving (i) (ii) }\)
\(\lambda_0=9 \lambda\)
\(\phi=\frac{\mathrm{hc}}{\lambda_0}\)
\(\mathrm{K}_{\max }=\mathrm{eV}_0\)
\(8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . \text { (i) }\)
\(2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . . \text { (ii) }\)
\(\text {on solving (i) (ii) }\)
\(\lambda_0=9 \lambda\)
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