JEE Advanced levelSingle correct optionDifficulty 3 of 3as041
Light of wavelength 200nm falls on a metal of work function 4.2eV. The fastest photoelectrons enter a uniform magnetic field of 1.0mT perpendicular to their velocity. Use h=6.626×10−34Js, c=3.0×108ms−1, me=9.1×10−31kg, e=1.6×10−19C. The radius of their circular path is closest to
A3.38 mm
B4.78 mm
C9.57 mm
D0.00 m
Answer: B
Photon energy = 6.212 eV, so Kmax = 6.212 - 4.2 = 2.012 eV = 3.219e-19 J. v = sqrt(2K/m) = 8.411e+05 m/s. r = mv/(eB) = (9.1e-31)(8.411e+05)/((1.6e-19)(1e-3)) = 4.784e-03 m = 4.78 mm.