Dual Nature of Matter

JEE Main level, single correct, medium

183 of 200

Question 183 of 200

Radiation of wavelength λ\lambda is incident on a photocell. The fastest emitted photoelectron has a speed vv. If the wavelength is changed to 3λ4\dfrac{3\lambda }{4}, the speed of the fastest emitted photoelectron will be

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Answer: (C) >v(43)12\gt v{\left(\dfrac{4}{3}\right)}^{\dfrac{1}{2}}

12mv2=hcλ−ϕ\dfrac{1}{2}{mv}^{2}=\dfrac{hc}{\lambda }-\phi ...(i)

12mv′2=43hcλ−ϕ\dfrac{1}{2}mv{'}^{2}=\dfrac{4}{3}\dfrac{hc}{\lambda }-\phi ...(ii)

From Eqn. (i)
hcλ=12mv2+ϕ\dfrac{hc}{\lambda }=\dfrac{1}{2}{mv}^{2}+\phi

On putting this Equation (ii)

12mv′2=43(12mv2+ϕ)−ϕ\dfrac{1}{2}{mv}^{'2}=\dfrac{4}{3}\left(\dfrac{1}{2}{mv}^{2}+\phi \right)-\phi

v′2=43v2+2ϕ3mv{'}^{2}=\dfrac{4}{3}{v}^{2}+\dfrac{2\phi }{3m}

⇒v′>43v\Rightarrow v'\gt \sqrt{\dfrac{4}{3}}v

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