JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
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 -
- A \(\frac{\lambda}{4}\)
- B \(4 \lambda\)
- C \(\frac{3}{2} \lambda\)
- D \(3 \lambda\)
Answer & Solution
Correct Answer
(D) \(3 \lambda\)
Step-by-step Solution
Detailed explanation
From the equation of photoelectric effect \(eV _0=\frac{ hc }{\lambda}-\phi_0=\frac{ hc }{\lambda}-\frac{ hc }{\lambda_0}\) \(\frac{ eV _0}{4}=\frac{ hc }{2 \lambda}=\frac{ hc }{\lambda_0}\)…
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