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Q.

An electron revolving in an orbit of radius 0.5A⁰ in a hydrogen atom executes 10¹⁶ revolutions per second. The magnetic moment of electron due to its orbital motion will be

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a

10^–13 A m²

b

653 × 10^–26 A m²

c

1.256 × 10^–23 A m²

d

256 × 10^–26 A m²

answer is A.

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Detailed Solution

$M = iA = \frac{e}{T} {πr}^{2} = {efπr}^{2}$
$∴ M = 1 .6 \times 10^{- 19} (10^{16}) \times 3 .14 {(0 .5 \times 10^{- 10})}^{2}$
$∴ M = 1 .256 \times 10^{- 23}$

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