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

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Question : 2 of 4
Marks: +1, -0
4. (a) Calculate the frequency of a photon of energy 6.5×10−19 J6.5 \times 10^{-19} \text{ J}.
(b) Can this photon cause emission of an electron from the surface of Cs of work function 2.14 eV2.14 \text{ eV} ? If yes, what will be maximum kinetic energy of the photoelectron?
3
Solution:  
(a) Frequency of photon =Eh= \frac{E}{h}
∴ Frequency =v=6.5×10−196.6×10−34=1015 Hz= v = \frac{6.5 \times 10^{-19}}{6.6 \times 10^{-34}} = 10^{15} \text{ Hz}
(b) work function of Cs=ϕ0=2.14 eV\mathrm{Cs} = \phi_0 = 2.14 \text{ eV}
Threshold frequency =v0=ϕ0h= v_0 = \frac{\phi_0}{h}
=2.14×1.6×10−196.6×10−34=0.5×1015 Hz= \frac{2.14 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} = 0.5 \times 10^{15} \text{ Hz}
Since the energy of incident photon is more than the threshold frequency, emission of photoelectrons will be possible.
KEmax=KE_{\text{max}} = Energy of incident photon −- Work function
∴KEmax=6.5×10−19−2.14×1.6×10−19\therefore KE_{\text{max}} = 6.5 \times 10^{-19} - 2.14 \times 1.6 \times 10^{-19}
=3.1×10−19 J= 3.1 \times 10^{-19} \text{ J}
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