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

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Question : 2 of 5
Marks: +1, -0
The work function of a metal is 2.31eV2.31 \mathrm{eV}. Photoelectric emission occurs when light of frequency 6.4×10146.4 \times 10^{14} Hz\mathrm{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= v = 6.4 \times 10^{14} \mathrm{Hz}
energy of incident radiation =E=Hv=6.6×10−34×6.4×1014=42.24×10−20J= E = Hv = 6.6 \times 10^{-34} \times 6.4 \times 10^{14} = 42.24 \times 10^{-20} \mathrm{J}
(ii) KEmax⁡=hv−ϕ0KE_{\max} = hv - \phi_0
∴KEmax⁡=42.24×10−20−2.31×1.6×10−19\therefore KE_{\max} = 42.24 \times 10^{-20} - 2.31 \times 1.6 \times 10^{-19}
=5.28×10−20J= 5.28 \times 10^{-20} \mathrm{J}
(iii) If stopping potential =Vs= V_s, then
eVs=KEmax⁡eV_s = KE_{\max}
∴Vs=KEmax⁡e\therefore V_s = \frac{KE_{\max}}{e}
=5.28×10−201.6×10−19= \frac{5.28 \times 10^{-20}}{1.6 \times 10^{-19}}
=3.3×10−1=0.33V= 3.3 \times 10^{-1} = 0.33 \mathrm{V}
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