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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2020
- Medium
- MCQ
When radiation of wavelength \(\lambda\) is used to illuminate a metallic surface, the stopping potential is \(V.\) When the same surface is illuminated with radiation of wavelength \(3 \lambda,\) the stopping potential is \(\frac{ V }{4} .\) If the threshold wavelength for the metallic surface is \(n \lambda\) then value of \(n\) will be......
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(9\)
Step-by-step Solution
Detailed explanation
\(\frac{ hc }{\lambda}=\frac{ hc }{\lambda_{0}}+ eV\)\(...(i)\)
\(\frac{ hc }{3 \lambda}=\frac{ hc }{\lambda_{0}}+\frac{ e \cdot V }{4}\)\(...(ii)\)
(multiply by \(4\))
\(\frac{4 hc }{3 \lambda}=\frac{4 hc }{\lambda_{0}}+ eV\)\(...(iii)\)
From \((i)\) \(\&\) \((iii)\)
\(\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}}=\frac{4 h c}{3 \lambda}-\frac{4 h c}{\lambda_{0}}\)
\(-\frac{ hc }{3 \lambda}=-\frac{3 hc }{\lambda_{0}}\)
\(9 \lambda=\lambda_{0}\)
\(n=9\)
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