JEE Main 26-Aug-2021 Shift 1 Solved Paper

By Einstein’s photoelectric equation, we have
⇒hf=ϕ+(KE)max
where, f is the frequency, ϕ is the work-function of metal and (KE)max is the maximum kinetic energy of emitted photoelectron.
⇒

hc

λ

=ϕ+eV0
⇒eV0=

hc

λ

−ϕ
Here, λ=280‌nm and ϕ=2.5‌eV
⇒e(V0)1=

1240

280

−2.5
⇒e(V0)1=1.93‌eV
⇒(V0)1=1.93V
Similarly, stopping potential for the light of wavelength of 400 nm,
e(V0)2=

1240

400

−2.5
⇒(V0)2=0.6V
∴ Change in stopping potential,
ΔV=(V0)1−(V0)2
=1.93−0.6=1.33≃1.3V