Dual Nature of Matter

JEE Main level, single correct, medium

128 of 200

Question 128 of 200

A metallic surface is illuminated with radiation of wavelength λ\lambda, the stopping potential is V0{V}_{0}. If the same surface is illuminated with radiation of wavelength 2λ2\lambda, the stopping potential becomes V04\dfrac{{V}_{0}}{4}. The threshold wavelength for this metallic surface will be

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Answer: (A) 3λ3\lambda

Let the threshold frequency be λ0{\lambda }_{0}.
By the equation of photoelectric effect, for wavelength λ\lambda,
eV0=hc(1λ−1λ0)...(i)e{V}_{0}=hc\left(\dfrac{1}{\lambda }-\dfrac{1}{{\lambda }_{0}}\right)...(i)
For the wavelength 2λ2\lambda,
eV04=hc(12λ−1λ0)...(ii)\dfrac{e{V}_{0}}{4}=hc\left(\dfrac{1}{2\lambda }-\dfrac{1}{{\lambda }_{0}}\right)...(ii)
Dividing (i) by (ii)
4=hc(λ0−λλλ0)hc(λ0−2λ2λλ0)⇒4=2(λ0−λ)λ0−2λ⇒2λ0=6λ⇒λ0=3λ4=\dfrac{hc(\dfrac{{\lambda }_{0}-\lambda }{\lambda {\lambda }_{0}})}{hc\left(\dfrac{{\lambda }_{0}-2\lambda }{2\lambda {\lambda }_{0}}\right)} \Rightarrow 4=\dfrac{2({\lambda }_{0}-\lambda )}{{\lambda }_{0}-2\lambda } \Rightarrow 2{\lambda }_{0}=6\lambda \Rightarrow {\lambda }_{0}=3\lambda

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