JEE Main Physics Class 12 Dual Nature of Matter and Radiation Part 1 Questions

According to question, there are two EM waves with different frequency,
B1=B0‌sin(π×107c)‌t
and B2=B0‌sin(2π×107c)‌t
To get maximum kinetic energy we take the photon with higher frequency
using, B=B0‌sin‌ω‌t and ω=2πv⇒v=ω2π
B1=B0‌sin(π×107c)‌t‌ ⇒v1=

107

2

×c
B2=B0‌sin(2π×107c)‌t⇒v2=107c
where cis speed of light c=3×108m∕s
Clearly, v2>v1
so KE of photoelectron will be maximum for photon of higher energy.
v2=107cHz
hv=ϕ+KEmax
energy of photon
Eph=hv=6.6×10−34×107×3×109
Ephpn=6.6×3 ×10−19J
=

6.6×3×10−19

1.6×10−19

eV=12.375eV
KEmax=Eph−ϕ
=12.375−4.7=7.675eV ≈7.7eV