JEE Main 9 Jan 2019 Shift 2 Solved Paper

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