AP EAPCET 06-Jul-2022 Shift 2 Paper

Given, E=E0sin‌[(1.57×107m−1)(x−ct)]......(i)
Work function, φ0=1.9‌eV=1.9×1.6×10−19J
Given, planck's constant, h=6.64×10−34J− s
We know that,
hv=φ0+eV0 (Photoelectric equation)......(ii)
Where, hv= incident energy of one photon
φ0= work function of metal
˙e= charge of an electron
V0= stopping potential
Now, from Eq. (i)
Clearly 2π∕λ=1.57×107
Now, from Eq. (ii)
‌hv=φ0+eV0
⇒‌

hc

λ

=φ0+eV0 (∵v=c∕λ)
⇒‌

hc

λ

−φ0=eV0
⇒4.98×10−19−3.04×10−19‌=eV0
⇒10−19‌

(4.98−3.04)

e

‌=V0
⇒V0=‌

1.94×10−19

1.6×10−19

‌‌(∵e=1.6×10−19C)
⇒V0=1.21V
Stopping potential =1.2V≃1.1V