91Ó°ÊÓ

The expression for the internal magnetic field in the hydrogen atom is given as follows,

$\mathit{B}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\mathbf{.}\frac{\mathbf{e}}{\mathbf{m}{\mathbf{c}}^{\mathbf{2}}{\mathbf{r}}^{\mathbf{3}}}\mathit{L}$

Here,${\mathrm{Î\mu }}_0$is the permittivity of the free space with value role="math" localid="1658138846877" $8.9×{10}^{-10}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2$,e is the charge on electron with value $1.6×{10}^{-19}C,m$, is the mass of the electron with value $9.1×{10}^{-31}\mathrm{kg}$,c is the speed of light with value $3×{10}^8\mathrm{m}/\mathrm{s},\mathrm{r}$is the Bohr’s radius with value $0.53×{10}^{-10}\mathrm{m}$, and $L=\mathrm{Ä\S }$which is Planck’s constant with value $1.05×{10}^{-34}\mathrm{J}.\mathrm{s}$.

Assume r=a that is Bohr’s radius, and $L=\mathrm{Ä\S }$.

Substitute the values in the expression for the internal magnetic field in the hydrogen atom.

$$\begin{gathered} B=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{e}{m_e{c}^2{a}^3}\mathrm{Ä\S } \\ =\frac{1}{4\mathrm{Ï€}\left(8.9×{10}^{-10}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)×{\displaystyle \frac{1\mathrm{N}.\mathrm{m}}{1\mathrm{J}}}}.\frac{\left(1.6×{10}^{-19}C\right)\left(1.05×{10}^{-34}J.s\right)}{\left(9.1×{10}^{-31}kg\right){\left(3×{10}^{8}m/s\right)}^2\left(0.53×{10}^{-10}m\right)} \\ =12\mathrm{C}.\mathrm{m}/\mathrm{s}×\frac{1T}{1C.m/s} \\ =12\mathrm{T} \end{gathered}$$

It is known that the strong Zeeman field is $B_{ext}>>10\mathrm{T}$and the weak Zeeman field is $B_{ext}≪10T$. So,$B_{ext}≫B_{int}and{B}_{ext}≪B_{int}$ .