Q. A light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV. If the same surface is illuminated by a light source of wavelength $\frac\lambda2$, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV)
A) 3 eV
B) 2 eV
C) 6 eV
D) 5 eV
Explanation
[Solution by Dr. Abubacker] $h\nu_1=K+\phi=3$ eV. Halving the wavelength doubles the photon energy: $h\nu_2=6$ eV, so $K_{max}=6-1=5$ eV. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.