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CBSE Class 12 Physics 2022 Term 2 Delhi Set 1 Solved Paper

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Question : 9 of 12
Marks: +1, -0
(a) Calculate the energy and momentum of a photon in a monochromatic beam of wavelength 331.5 nm331.5\text{ nm}.3
(b) How fast should a hydrogen atom travel in order to have the same momentum as that of the photon in part (a)?
Solution:  
(a) Energy =E=hcλ= E = \frac{hc}{\lambda}
Or, E=6.6×10−34×3×108331.5×10−9E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{331.5 \times 10^{-9}}
∴E=6×10−19 J\therefore E = 6 \times 10^{-19}\text{ J}
Momentum =p=hλ= p = \frac{h}{\lambda}
Or, p=6.6×10−34331.5×10−9p = \frac{6.6 \times 10^{-34}}{331.5 \times 10^{-9}}
∴p=2×10−27 kg.m/s\therefore p = 2 \times 10^{-27}\text{ kg.m/s}
(b) Momentum of Hydrogen atom
=p=2×10−27 kg.m/s= p = 2 \times 10^{-27}\text{ kg.m/s}
p=mvp = mv
Or, v=pmv = \frac{p}{m}
Or, v=2×1027 kg.m/s1 a.m.u.v = \frac{2 \times 10^{27}\text{ kg.m/s}}{1\text{ a.m.u.}}
Or, v=2×10−27 kg.m/s1.66×10−27 kgv = \frac{2 \times 10^{-27}\text{ kg.m/s}}{1.66 \times 10^{-27}\text{ kg}}
∴v=1.2 m/s\therefore v = 1.2\text{ m/s}
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