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

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Question : 9 of 12
Marks: +1, -0
An electron is accelerated from rest through a potential difference of 100 V100\ \mathrm{V}.
Find:
(i) the wavelength associated with
(ii) the momentum and
(iii) the velocity required by the electron. 3
Solution:  
(i) Wavelength =λ=h2meV= \lambda = \frac{h}{\sqrt{2 m e V}}
∴λ=6.6×10−342×9.1×1031×1.6×10−19×100\therefore \lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{31} \times 1.6 \times 10^{-19} \times 100}}
=1.2×10−10 m=1.2 A˚= 1.2 \times 10^{-10}\ \mathrm{m} = 1.2\ \text{\AA}
(ii) Momentum =p=hλ= p = \frac{h}{\lambda}
∴p=6.6×10−341.2×10−10\therefore p = \frac{6.6 \times 10^{-34}}{1.2 \times 10^{-10}}
=5.5×10−24 kg m/s= 5.5 \times 10^{-24}\ \mathrm{kg\ m/s}
(iii) Momentum = p=mv
∴\therefore Velocity =v=pm= v = \frac{p}{m}
=5.5×10−249.1×10−31= \frac{5.5 \times 10^{-24}}{9.1 \times 10^{-31}}
=6×106 m/s= 6 \times 10^6\ \mathrm{m/s}
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