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In a hydrogen atom, the electron is making $6.6 \times 10^{15} \mathrm{revs}^{-1}$ around the nucleus in an orbit of radius $0.528 \mathrm{~A}$. The magnetic moment $\left(\mathrm{A}-\mathrm{m}^{2}\right)$ will be
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The correct answer is:
$1 \times 10^{-23}$
Current, $I=6.6 \times 10^{15} \times 1.6 \times 10^{-19}$
$=10.5 \times 10^{-4} \mathrm{~A} .$
Area $A=\pi R^{2}=3.142 \times(0.528)^{2} \times 10^{-20} \mathrm{~m}^{2}$
So, Magnetic moment $M=I A=10.5 \times 10^{-4} \times 3.142$
$\begin{aligned} \times(0.528)^{2} \times 10^{-20} & \\=& 10 \times 10^{-24}=10^{-23} \text { units } \end{aligned}$
$=10.5 \times 10^{-4} \mathrm{~A} .$
Area $A=\pi R^{2}=3.142 \times(0.528)^{2} \times 10^{-20} \mathrm{~m}^{2}$
So, Magnetic moment $M=I A=10.5 \times 10^{-4} \times 3.142$
$\begin{aligned} \times(0.528)^{2} \times 10^{-20} & \\=& 10 \times 10^{-24}=10^{-23} \text { units } \end{aligned}$
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