TS EAMCET · Physics · Dual Nature of Matter
When radiation of the wavelength \(\lambda\) is incident on a metallic surface, the stopping potential is \(4.8 \mathrm{~V}\). If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes \(1.6 \mathrm{~V}\). Then, the threshold wavelength for the surface is :
- A \(2 \lambda\)
- B \(4 \lambda\)
- C \(6 \lambda\)
- D \(8 \lambda\)
Answer & Solution
Correct Answer
(B) \(4 \lambda\)
Step-by-step Solution
Detailed explanation
Stopping potential. \(V_0=\frac{h c}{e}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\) where, \(V_0=\) stopping potential Ist case, \(4.8=\frac{h c}{e}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\) IInd case,…
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