Question

Derive an expression for the magnetic dipole moment of a revolving electron.

Answer 1

Hint: In this question,we apply the concept of dipole moment and concept of orbital motion.

Complete step by step answer:
As indicated by Bohr's atom model, the negatively charged electron is rotating around a positively charged nucleus in a circular orbit of radius. The spinning electron in a closed path establishes an electric flow. The movement of the electron in an anticlockwise way produces current in the clockwise direction.
Currently,\[I = \dfrac{e}{T}\], where T is the period of revolution of the electron.
If v is the orbital velocity of the electron, then
\[T = \dfrac{{2\pi r}}{v}\]
\[\therefore \]\[i = \dfrac{{ev}}{{2\pi r}}\]
Due to the orbital motion of the electron, there will be orbital magnetic moment \[\mathop \mu \nolimits_i \]
\[\mathop \mu \nolimits_i = iA\] Where A is the area of the orbit
\[\begin{gathered}
\mathop \mu \nolimits_i = \dfrac{{ev}}{{2\pi r}} \cdot \pi {r^2} \\
\mathop \mu \nolimits_i = \dfrac{{evr}}{2} \\
\end{gathered} \]
If m is the mass of the electron,
Multiply denominator and numerator with m
 \[\mathop \mu \nolimits_i = \dfrac{e}{{2m}} \cdot \left( {mvr} \right)\]
Mr is the angular momentum \[\left( L \right)\] of the electron about the central nucleus.
\[\mathop \mu \nolimits_i = \dfrac{e}{{2m}} \cdot L\]- - - - - - - - - - - - - - - \[\left( 1 \right)\]
\[\dfrac{{\mathop \mu \nolimits_i }}{L} = \dfrac{e}{{2m}}\]
Where \[\dfrac{{\mathop \mu \nolimits_i }}{L}\] is the gyromagnetic ratio and it is constant. Its value is \[8.8{\text{ }} \times {\text{ }}{10^{10\;}}C{\text{ }}k{g^{ - 1}}\]Bohr hypothesized that the angular momentum has only discrete set of values given by the equation.
\[L = \dfrac{{nh}}{{2\pi }}\] - - - - - - - - - - - - - - - \[\left( 2 \right)\]
Where n is a natural number and h is the Planck’s constant \[ = 6.626{\text{ }} \times {\text{ }}{10^{ - 34}}\;Js.\]
Substituting equation \[\left( 2 \right)\;\] in equation \[\left( 1 \right)\]
\[\mathop \mu \nolimits_i = \dfrac{e}{{2m}} \cdot \dfrac{{nh}}{{2\pi }}\]
\[ = \dfrac{{neh}}{{4\pi m}}\] - - - - - - - - - - - \[\left( 3 \right)\]
The minimum value of magnetic moment is
\[{\left( {\mathop \mu \nolimits_i } \right)_{\min }} = \dfrac{{eh}}{{4\pi m}}\] , Where \[n{\text{ }} = {\text{ }}1\]
The value of \[\dfrac{{eh}}{{4\pi m}}\]is called Bohr magneton.
By substituting the values of \[e,{\text{ }}h\;\] and \[m,\] the value of Bohr magneton is found to be \[9.27{\text{ }} \times {\text{ }}{10^{-24}}\;A{m^2}\].

Note:Always remember the exact value of Bohr magneton because it is useful in solving various types of numerical problems and also keep proper knowledge of orbital motion.