Dual Nature of Matter

JEE Main level, single correct, medium

121 of 200

Question 121 of 200

When a metal surface is illuminated by light of wavelength λ\lambda, the stopping potential is 8V8V. When the same surface is illuminated by light of wavelength 3λ3\lambda, stopping potential is 2V2V. The threshold wavelength for this surface is :

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Answer: (C) 9λ9\lambda

From photoelectric effect equation, we can write E=ϕ+KmaxE=\phi +{K}_{max}
and work function can be written as ϕ=hcλ0\phi =\dfrac{hc}{{\lambda }_{0}}.
Also, maximum kinetic energy Kmax=eV0{K}_{max}=e{V}_{0}
For first case:
8e=hcλ−hcλ08e=\dfrac{hc}{\lambda }-\dfrac{hc}{{\lambda }_{0}} ... (i)
For second case:
2e=hc3λ−hcλ02e=\dfrac{hc}{3\lambda }-\dfrac{hc}{{\lambda }_{0}} ... (ii)
After using equation (ii) in (i), we get
4(hc3λ−hcλ0)=hcλ−hcλ04\left(\dfrac{hc}{3\lambda }-\dfrac{hc}{{\lambda }_{0}}\right)=\dfrac{hc}{\lambda }-\dfrac{hc}{{\lambda }_{0}}
⇒λ0=9λ\Rightarrow {\lambda }_{0}=9\lambda

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