JEE Main 2016 · 10 Apr · Shift 1 · Physics Q21

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PhysicsSingle correct+4 / -1Question 21 of 90

A photoelectric surface is illuminated successively by monochromatic light of wavelengths λ and $\dfrac\lambda2$ . 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 :

\dfrac{hc}{3\lambda}

\dfrac{hc}{2\lambda}

\dfrac{hc}\lambda

\dfrac{3hc}\lambda

Answer and worked solution

Official answer: B

$K_1=\frac{hc}\lambda-\phi$ and $K_2=\frac{2hc}\lambda-\phi=3K_1$ . So $2\phi=\frac{hc}\lambda$ and $\phi=\frac{hc}{2\lambda}$ .

Topics:Photoelectric effect

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JEE Main 2016 · 10 Apr · Shift 1, Physics question 21