Dual Nature of Matter

JEE Main level, single correct, medium

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Question 1 of 200

When radiation of wavelength λ\lambda falls on a metal, stopping potential is 3V03V_0. When wavelength is 2λ2\lambda, stopping potential is V0V_0. The threshold wavelength is:

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Answer: (D) 4λ4\lambda

e(3V0)=hcλ−ϕe(3V_0) = \frac{hc}{\lambda} - \phi and e(V0)=hc2λ−ϕe(V_0) = \frac{hc}{2\lambda} - \phi. Subtracting 3e(V0)−e(3V0)=0  ⟹  3(hc2λ−ϕ)=hcλ−ϕ  ⟹  2ϕ=hc2λ  ⟹  ϕ=hc4λ  ⟹  λ0=4λ3e(V_0) - e(3V_0) = 0 \implies 3\left(\frac{hc}{2\lambda} - \phi\right) = \frac{hc}{\lambda} - \phi \implies 2\phi = \frac{hc}{2\lambda} \implies \phi = \frac{hc}{4\lambda} \implies \lambda_0 = 4\lambda.

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