The magnetic field associated with a light wave is given, at the origin, by B=B0[sin(3.14×10⁷ct)+sin(6.28×10⁷ct)]. If this light falls on a silver plate having a work function of 4.7eV, what will be the maximum kinetic energy of the photo electrons?

B=B0sin(π×10⁷Ct)+B0sin(2π×10⁷Ct)
since there are two EM waves with different frequency, to get maximum kinetic energy we take the photon with higher frequency
B1=B0sin(π×10⁷Ct)
v1=10⁷C/2
B2=B0sin(2π×10⁷Ct)
v2=10⁷C
Where C is speed of light
C=3×10⁸m/s
v2>v1
so KE of photoelectron will be maximum for photon of higher energy.
v2=10⁷C
E=hv=Φ+KEmax
energy of photon
Eph=hv=6.6×10⁻³⁴×10⁷×3×10⁸
eph=6.6×3×10⁻¹⁹J
=6.6×3×10⁻¹⁹/1.6×10⁻¹⁹eV
=12.375eV
KEmax=Eph−Φ=12.375−4.7=7.675eV≈7.72eV