A photoelectric surface is illuminated successively by monochromatic light of wavelengths λ and(λ)/(2) . If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case

Correct answer: A. (hc)/(2λ) K . E _ λ = (hc)/(λ) - ϕ K . E _ (λ)/(2) = (hc)/((λ)/(2)) - ϕ K . E _ (λ)/(2) = 3(K . E _ λ ) ⇒ ϕ = (hc)/(2λ)

JEE Main 201610 AprilOnline

Exam: JEE Main
Year: 2016
Date: 10 April
Mode: Online
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2016 · 10 April · Online · Physics · Dual Nature of Matter

A photoelectric surface is illuminated successively by...

A photoelectric surface is illuminated successively by monochromatic light of wavelengths

\lambda and\dfrac{\lambda}{2} .

If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :

Options

A $\dfrac{hc}{2\lambda}$
B $\dfrac{hc}{\lambda}$
C $\dfrac{hc}{3\lambda}$
D $\dfrac{3hc}{\lambda}$

Correct answer

A. $\dfrac{hc}{2\lambda}$

Step-by-step solution

K . {E}_{\lambda} = \dfrac{hc}{\lambda} - ϕ

K . {E}_{\dfrac{\lambda}{2}} = \dfrac{hc}{\dfrac{\lambda}{2}} - ϕ

K . {E}_{\dfrac{\lambda}{2}} = 3\left(K . {E}_{\lambda}\right)

⇒ ϕ = \dfrac{hc}{2\lambda}

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