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JEE MAIN 2020
02-09-20 S1
Question
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 $\frac{\mathrm{V}}{4}$. If the threshold wavelength for the metallic surface is $n \lambda$ then value of $n$ will be $\_\_\_\_$ .
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Question Tags
JEE Main
Physics
Medium
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