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

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Question : 4 of 4
Marks: +1, -0
(a) Give an example each of a metal from which photoelectric emission takes place when irradiated by (i) UV light (ii) visible light. 3
(b) The work function of a metal is 4.50eV4.50 \text{eV}. Find the frequency of light to be used to eject electrons from the metal surface with a maximum kinetic energy of 6.06×10−19J6.06 \times 10^{-19} \text{J}. 3
Solution:  
(a) (i) Photoelectric emission takes place when Zn\mathrm{Zn} is irradiated with UV radiation.
(ii) Photoelectric emission takes place when Na\mathrm{Na} is irradiated with visible light.(b) Maximum kinetic energy of photoelectron
== Energy of incident photon - Work function
Or, 6.06×10−19=hv−4.5×1.6×10196.06\times10^{-19}=hv-4.5\times1.6\times10^{19}
Or, hv=13.26×10−19hv = 13.26\times10^{-19}
Or, v=13.26×10−196.6×10−34v=\frac{13.26\times10^{-19}}{6.6\times10^{-34}}
∴v=2×1015Hz\therefore v=2\times10^{15} \text{Hz}
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