CBSE Class 12 Physics 2022 Term 2 Outside Delhi Set 3 Solved Paper

Question : 2

Total: 5

The work function of a metal is 2.31‌eV. Photoelectric emission occurs when light of frequency 6.4×1014 Hz is incident on the metal surface. Calculate (i) the energy of the incident radiation, (ii) the maximum kinetic energy of the emitted electron and (iii) the stopping potential of the surface.
3

Solution:

(i) Frequency of incident radiation =v=6.4×1014Hz
energy of incident radiation =E=Hv=6.6×10−34×6.4×1014=42.24×10−20J
(ii) KEmax=hv−ϕ0
∴KEmax=42.24×10−20−2.31×1.6×10−19
=5.28×10−20J
(iii) If stopping potential =Vs, then
eVs=KEmax
∴Vs=

KEmax

5.28×10−20

1.6×10−19

=3.3×10−1=0.33V

Go to Question:

Prev Question

Next Question