B = B0 sin (π × 107 C)t + B0 sin (2π × 107 C)tsince there are two EM waves with differentfrequency, to get maximum kinetic energy wetake the photon with higher frequency B1 = B0 sin (π × 107C) t , v1 =
107
2
× C B2 = B0 sin (2π × 107 C) t , v2 = 107 C where C is speed of light C = 3 × 108 m/s v2 > v1 so KE of photoelectron will be maximum forphoton of higher energy v2 = 107 C Hz hv = Φ + KEmax energy of photon Eph = hv = 6.6 × 10−34 × 107 × 3 × 109 Eph = 6.6 × 3 × 10−19 J =
6.6×3×10−19
1.6×10−19
eV = 12.37 eV KEmax = Eph - Φ = 12.375 - 4.7 = 7.675 eV ~ 7.7 eV