What are the dimensions of $\chi$,the magnetic susceptibility? Consider an $H$-atom. Guess an expression for $\chi$,up to a constant,by constructing a quantity of dimensions of $\chi$ out of parameters of the atom: $e, m, v, R$ and $\mu_0$. Here,$m$ is the electronic mass,$v$ is electronic velocity,$R$ is Bohr radius. Estimate the number so obtained and compare with the value of $|\chi| \approx 10^{-5}$ for many solid materials.

(A) Magnetic susceptibility $\chi$ is defined as the ratio of magnetization $M$ to magnetic field intensity $H$. Since both $M$ and $H$ have the same units $(A/m)$,$\chi$ is a dimensionless quantity,i.e.,$[M^0 L^0 T^0]$.
To construct an expression for $\chi$ using $e, m, v, R, \mu_0$,we note that $\mu_0$ has dimensions $[M L T^{-2} Q^{-2}]$ and $e^2$ has dimensions $[Q^2]$. Thus,$\mu_0 e^2$ has dimensions $[M L T^{-2}]$.
Let $\chi = \mu_0 e^2 m^a v^b R^c$.
Substituting dimensions: $[M^0 L^0 T^0] = [M L T^{-2}] [M]^a [L T^{-1}]^b [L]^c$.
Equating powers of $M, L, T$:
For $M$: $1 + a = 0 \implies a = -1$.
For $T$: $-2 - b = 0 \implies b = -2$.
For $L$: $1 + b + c = 0 \implies 1 - 2 + c = 0 \implies c = 1$.
Thus,$\chi \propto \frac{\mu_0 e^2 R}{m v^2}$.
Using Bohr's model,$v^2 = \frac{e^2}{4 \pi \epsilon_0 m R}$,so $\frac{e^2}{m R} \approx v^2$. Substituting this,$\chi \approx \mu_0 \epsilon_0 v^2 \approx \frac{v^2}{c^2}$.
Given $v \approx \alpha c$ (where $\alpha \approx 1/137$),$\chi \approx \alpha^2 \approx (1/137)^2 \approx 5 \times 10^{-5}$,which is consistent with the order of $10^{-5}$ for diamagnetic materials.