Question

An electron in an atom revolves around the nucleus in an orbit of radius $0.53\stackrel{0}{A}$. Calculate the equivalent magnetic moment, if the frequency of revolution of electron is $6.8\times {10}^9$ $MHz$(Given :$e=1.6\times {10}^{-19}$ ).

Answer 1

The formula of magnetic moment is $M=NIA$

where $N$=no. of turns=$1$

$I$=Current,

$A$=Area

$M=NIA$

$=IA\Rightarrow \frac{q}{t}A=\Rightarrow =Qfa(\frac{1}{t}=frequency)$

$=1.6\times {10}^{-19}\times 6.8\times {10}^9\times {10}^6\times \pi \times 0.53\times {10}^{-10}(1A^0={10}^{-10})$

$=5.766\times \pi \times {10}^{-19+9+6-10}$

$=18.10\times {10}^{-14}A-m^2$