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In question $4.11$ obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron?
Solution:
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Verified Answer
We know,
$$
\begin{aligned}
&\omega_{\mathrm{c}}=\frac{\mathrm{qB}}{\mathrm{m}} \Rightarrow 2 \pi v=\frac{\mathrm{qB}}{\mathrm{m}} \Rightarrow v=\frac{\mathrm{qB}}{2 \pi \mathrm{m}} \\
&=\frac{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}{2 \times 3.14 \times 9.1 \times 10^{-31}}=0.182 \times 10^8=18.2 \times 10^6 \\
&=18.2 \mathrm{MHz} .
\end{aligned}
$$
Answer does not depend on the speed of the electron.
$$
\begin{aligned}
&\omega_{\mathrm{c}}=\frac{\mathrm{qB}}{\mathrm{m}} \Rightarrow 2 \pi v=\frac{\mathrm{qB}}{\mathrm{m}} \Rightarrow v=\frac{\mathrm{qB}}{2 \pi \mathrm{m}} \\
&=\frac{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}{2 \times 3.14 \times 9.1 \times 10^{-31}}=0.182 \times 10^8=18.2 \times 10^6 \\
&=18.2 \mathrm{MHz} .
\end{aligned}
$$
Answer does not depend on the speed of the electron.
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