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← JEE Advanced 2016 · Paper 2
PhysicsMultiple correct+4 / -2Question 8 of 54

JEE Advanced 2016 · Paper 2, Physics question 8

Light of wavelength λph\lambda_{ph} falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is ϕ\phi and the anode is a wire mesh of conducting material kept at a distance dd from the cathode. A potential difference VV is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is λe\lambda_e, which of the following statement(s) is(are) true?

An evacuated tube with a cathode plate on the left and a wire-mesh anode on the right; light (wavy arrow) falls on the cathode, and electrons travel to the mesh and pass through it. A battery of voltage V (negative to the cathode, positive to the anode) is connected across the electrodes.
Choose one or more options
Answer and worked solution

Official answer: C

Fastest electrons. They leave the cathode with hcλph−ϕ\frac{hc}{\lambda_{ph}}-\phi and gain eVeV, so at the anode
K=hcλph−ϕ+eV,λe=h2mK.K=\frac{hc}{\lambda_{ph}}-\phi+eV,\qquad\lambda_e=\frac h{\sqrt{2mK}}.

  • (A) Increasing ϕ\phi or λph\lambda_{ph} lowers KK, so λe\lambda_e increases. False.
  • (B) KK does not depend on dd. False.
  • (C) For eV≫ϕeV\gg\phi, K≈eVK\approx eV and λe∝V−1/2\lambda_e\propto V^{-1/2}. Making VV four times halves λe\lambda_e. True.
  • (D) λe\lambda_e is not proportional to λph\lambda_{ph}. False.

(C)