JEE Advanced levelOne or more correct optionsDifficulty 3 of 3as053
Light of wavelength 365nm is incident on metals A and B of work functions 2.0eV and 3.0eV. Use h=6.626×10−34Js, c=3.0×108ms−1, 1eV=1.6×10−19J. Select all correct statements.
AThe stopping potential for B is about 0.40 V
BThe ratio of stopping potentials, A : B, is 2 : 1
CBoth metals emit photoelectrons
DThe maximum kinetic energy of electrons from A is about 1.40 eV
Answer: A, C, D
Photon energy = hc/λ = 3.404 eV, above both work functions, so both emit. Kmax(A) = 3.404 - 2.0 = 1.40 eV. Kmax(B) = 3.404 - 3.0 = 0.40 eV, so V0(B) = 0.40 V. Ratio of stopping potentials A:B = 1.40:0.40 = 3.5:1, not 2:1.