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

5 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.

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

4.12 Let the frequency of revolution = <math> <mi>ν</mi> </math> Angular frequency = <math> <mi>ω</mi> <mo>=</mo> <mn>2</mn> <mi>π</mi> <mi>ν</mi> </math> Now, velocity of electron, v = r <math> <mi> </mi> <mi>ω</mi> </math> Since in circular orbit, magnetic force is balanced by the centripetal force, we can writeevB <math> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>m</mi> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> </math> or eB = <math> <mfrac> <mrow> <mrow> <mi>m</mi> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mi>m</mi> <mo> (</mo> <mi mathvariant="normal">r</mi> <mi> </mi> <mi>ω</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>π</mi> <mi>ν</mi> <mi>m</mi> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> </math> This frequency is independent of the speed of electron. <math> <mi>ν</mi> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mn>1.6</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>6.5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>4</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>π</mi> <mo>×</mo> <mn>9.1</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>31</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>18.19</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </math> Hz = 18.19 MHz

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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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