MHT CET · Physics · Dual Nature of Matter
When radiations of wavelength ' \(\lambda\) ' is incident on a metallic surface the stopping potential required is 4.8 volt. If same surface is illuminated with radiations of double the wavelength, then required stopping potential becomes 1.6 volt. Then the value of threshold wavelength for the surface is
- A \(2 \lambda\)
- B \(4 \lambda\)
- C \(6 \lambda\)
- D \(8 \lambda\)
Answer & Solution
Correct Answer
(C) \(6 \lambda\)
Step-by-step Solution
Detailed explanation
We know, \(\mathrm{V}=\frac{\mathrm{hc}}{\lambda}-\phi_0\)
\(\Rightarrow 4.8=\frac{\mathrm{hc}}{\lambda}-\phi\) ....(i)
Given \(\lambda=2 \lambda\)
\(\Rightarrow 1.6=\frac{h c}{2 \lambda}-\phi\) ....(ii)
\(\therefore \quad\) Dividing (i) by (ii),
\(3\left(\frac{\mathrm{hc}}{2 \lambda}-\phi\right)=\frac{\mathrm{hc}}{\lambda}-\phi\)
\(3 \frac{\mathrm{hc}}{2 \lambda}-3 \phi=\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)\)
\(\frac{\mathrm{hc}}{2 \lambda}=2 \phi\)
\(\therefore \quad\) Threshold wavelength is \(4 \lambda\)
\(\Rightarrow 4.8=\frac{\mathrm{hc}}{\lambda}-\phi\) ....(i)
Given \(\lambda=2 \lambda\)
\(\Rightarrow 1.6=\frac{h c}{2 \lambda}-\phi\) ....(ii)
\(\therefore \quad\) Dividing (i) by (ii),
\(3\left(\frac{\mathrm{hc}}{2 \lambda}-\phi\right)=\frac{\mathrm{hc}}{\lambda}-\phi\)
\(3 \frac{\mathrm{hc}}{2 \lambda}-3 \phi=\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)\)
\(\frac{\mathrm{hc}}{2 \lambda}=2 \phi\)
\(\therefore \quad\) Threshold wavelength is \(4 \lambda\)
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