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

Question : 2

Total: 4

4. (a) Calculate the frequency of a photon of energy 6.5×10−19J.
(b) Can this photon cause emission of an electron from the surface of Cs of work function 2.14‌eV ? If yes, what will be maximum kinetic energy of the photoelectron?
3

Solution:

(a) Frequency of photon =

∴ Frequency =v=

6.5×10−19

6.6×10−34

=1015Hz
(b) work function of Cs=ϕ0=2.14eV
Threshold frequency =v0=

ϕ0

2.14×1.6×10−19

6.6×10−34

=0.5×1015Hz
Since the energy of incident photon is more than the threshold frequency, emission of photoelectrons will be possible.
KEmax= Energy of incident photon − Work function
∴KEmax=6.5×10−19−2.14×1.6×10−19
=3.1×10−19J

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