Dual Nature of Matter

JEE Main level, single correct, medium

153 of 200

Question 153 of 200

In a photoelectric experiment, ultraviolet light of wavelength 280nm280nm is used with lithium cathode having work function ϕ=2.5eV\phi =2.5eV. If the wavelength of incident light is switched to 400nm400nm, find out the change in the stopping potential.
(h=6.63×10−34Js,c=3×108ms−1)\left(h=6.63\times {10}^{-34}Js,c=3\times {10}^{8}{ms}^{-1}\right)

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Answer: (C) 1.3V1.3V

From the photoelectric equation,
eV0=hcλ−ϕe{V}_{0}=\dfrac{hc}{\lambda }-\phi
Using this equation for the given two cases,
eV1=hc280nm−ϕe{V}_{1}=\dfrac{hc}{280nm}-\phi
eV2=hc400nm−ϕe{V}_{2}=\dfrac{hc}{400nm}-\phi
On subtracting,
e(V1−V2)=1240(120280×400)=1.3eVe\left({V}_{1}-{V}_{2}\right)=1240\left(\dfrac{120}{280\times 400}\right) =1.3eV
⇒(V1−V2)=1.3V\Rightarrow \left({V}_{1}-{V}_{2}\right)=1.3V

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