Question

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

Answer 1

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

\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.