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Q.

An electron revolving in an orbit of radius $0.5 \overset{0}{A}$ in a hydrogen atom executes 10 revolutions per second. The magnetic moment of electron due to its orbital motion will be

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a

$256 \times 10^{- 26} {Am}^{2}$

b

$653 \times 10^{- 41} {Am}^{2}$

c

$1256 \times 10^{- 41} {Am}^{2}$

d

Zero

answer is A.

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Detailed Solution

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$\begin{array}{l} \frac{M}{L} = \frac{q}{2 M} \\ M = (\frac{q}{2 M}) L \end{array}$

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