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Q)

Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depends on the speed of the electron ? Explain

$\begin{array}{1 1} 18.18 \times 10^6\; Hz \\ 18.18 \times 10^7 Hz \\ 1.818 \times 10^6 Hz \\ 1.818 \times 10^5 Hz \end{array} $

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A)
Solution :
Given $B= 6.5 \;G$
$\qquad= 6.5 \times 10^{-4}T$
$v=4.8 \times 10^6\;m/s$
$e= 1.6 \times 10^{-19}$
$m_e= 9.1 \times 10 ^{-31}$
$\large\frac{mv^2}{r}$$=qvB$
=> $\large\frac{mv}{r}$$=qB$
=> $w =\large\frac{qB}{m}$
$2 \pi n =\large\frac{qB}{m}$
$n= \large\frac{qB}{2 \pi m}$
Frequency of revolution of electron in the orbit.
$v= \large\frac{B_q}{2 \pi m_e}$
$\quad=\large\frac{B_e}{2 \pi m}$
$\quad= \large\frac{6.5 \times 10^{-4} \times 1.6 \times 10^{-3}}{2 \times 3.14 \times 9.1 \times 10^{-3}}$
$\quad= 18.18 \times 10^6 HZ$
Answer : $ 18.18 \times 10^6\; Hz$
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