JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2023
- Medium
- MCQ
A metallic surface is illuminated with radiation of wavelength \(\lambda\), the stopping potential is \(V_0\). If the same surface is illuminated with radiation of wavelength \(2 \lambda\), the stopping potential becomes \(\frac{V_0}{4}\). The threshold wavelength for this metallic surface will be -
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(3 \lambda\)
Step-by-step Solution
Detailed explanation
From the equation of photoelectric effect
\(eV _0=\frac{ hc }{\lambda}-\phi_0=\frac{ hc }{\lambda}-\frac{ hc }{\lambda_0}\)
\(\frac{ eV _0}{4}=\frac{ hc }{2 \lambda}=\frac{ hc }{\lambda_0}\)
\(\Rightarrow \frac{1}{4}\left(\frac{ hc }{\lambda}-\frac{ hc }{\lambda_0}\right)=\frac{ hc }{2 \lambda}-\frac{ hc }{\lambda_0}\)
\(\frac{1}{\lambda_0}-\frac{1}{4 \lambda_0}=\frac{1}{2 \lambda}-\frac{1}{4 \lambda}\)
\(\frac{3}{4 \lambda_0}=\frac{1}{4 \lambda}\)
\(\Rightarrow \lambda_0=3 \lambda\)
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