An electron is revolving with speed $v$ in a circular orbit of radius r. Obtain the expression of gyromagnetic ratio. What is a Bohr magneton? Write its value.

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Solution

Consider electron $e$ is revolving around nucleus with speed $v$ .
Period of revolution $= \frac{c i r c u m f e r e n c e}{v e l o c i t y}$
$T = \frac{2 π r}{v}$ ---- 1
Circulating current $I = \frac{e}{T}$ ----- 2
$= \frac{e}{2 π r / v} = \frac{e v}{2 π r}$ ---- 3
Magnitude of magnetic moment associated with the circular current is:
$M_{0} = I A = \frac{e v}{2 π r} \times π r^{2}$ ---- 4
$= \frac{e v r}{2}$ ------- 5
Multiplying and dividing the right-hand side of equation (5) by the mass of
electron, $m_{e}$ then,
$M_{0} = \frac{e}{2 m_{e}} (m_{e} v r)$ ---- 6
$= \frac{e}{2 m_{e}} \times L_{0}$ ---- 7
Here $L_{0} = m_{e}$ , $v r =$ angular momentum of the electron revolving around the nucleus.
The ratio of magnetic dipole moment with angular momentum of
revolving electron is called the gyromagnetic ratio.

Gyrometric ratio $= \frac{M_{0}}{L_{0}} = \frac{e}{2 m_{e}} = 8.8 \times 10^{10} C / k g$

Bohr magneton (symbol $μ B$ ) is a physical constant and the natural unit for expressing the magnetic moment of an electron caused by either its orbital or spin angular momentum

The Bohr magneton is defined in SI units by

$μ_{B} = \frac{e ℏ}{2 m_{e}}$

$μ_{B} = 9.274 \times 10^{- 24} J T^{- 1}$

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