First slide

The bar magnet

Question

In hydrogen atom the electron is making 6.6 x 10¹⁵rev/sec. around the nucleus in an orbit of radius o .528 $\overset{o}{A}$ . The magnetic moment (Am²) will be

Easy

A

1 x 10^-15
B

1 x 10^-10
C

1 x 10^-23
D

1 x 10^-27

Solution

Given that, v = 6.6 x 10¹⁵ rev./sec.
and r = 0.528 $\overset{o}{A}$ = 0.528 x 10^-10m
We know that, M = i A = (e/T) x A
$\begin{array}{l} ∴ M = (ev) \times {πr}^{2} \\ = (1 \cdot 6 \times 10^{- 19}) (6 \cdot 6 \times 10^{15}) \times 3 \cdot 14 \times {(0 \cdot 528 \times 10^{- 10})}^{2} \\ = 0 \cdot 925 \times 10^{- 23} {Am}^{2} \end{array}$

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