4.12 In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

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Payal Gupta | Contributor-Level 10

8 months ago

4.12 Let the frequency of revolution = $ν$

Angular frequency = $ω = 2 π ν$

Now, velocity of electron, v = r $ω$

Since in circular orbit, magnetic force is balanced by the centripetal force, we can write

evB $= \frac{m v^{2}}{r}$ or eB = $\frac{m v}{r}$ = $\frac{m (r ω)}{r}$ = $\frac{2 π ν m r}{r}$

This frequency is independent of the speed of electron.

$ν = \frac{1.6 * 10^{- 19} * 6.5 * 10^{- 4}}{2 π * 9.1 * 10^{- 31}} = 18.19 * 10^{6}$ Hz = 18.19 MHz

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Physics Ncert Solutions Class 12th 2023

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