If the electron were a classical solid sphere, with radius

${\mathbf{r}}_{\mathbf{c}}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }\dot{{\mathbf{O}}_{\mathbf{0}}}{\mathbf{mc}}^{\mathbf{2}}}$

(the so-called classical electron radius, obtained by assuming the electron's mass is attributable to energy stored in its electric field, via the Einstein formula $\mathbf{E}\mathbf{=}{\mathbf{mc}}^{\mathbf{2}}$), and its angular momentum is $\left(1/2\right)\mathbf{h}$ then how fast (in $\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$) would a point on the "equator" be moving? Does this model make sense? (Actually, the radius of the electron is known experimentally to be much less than$\mathbf{5}\mathbf{.}\mathbf{156}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}{\mathbf{r}}_{\mathbf{c}}$, but this only makes matters worse).

In practise, spin is computed as a dimensionless spin quantum number by dividing the spin angular momentum by the decreased Planck constant, which has the same dimensions as angular momentum.

Angular momentum is $\mathrm{L}=\mathrm{I\omega }.$

where $I$is the moment of inertia, and for a sphere, it is equal to $2/5{\mathrm{mr}}^2$

role="math" localid="1655966871442" $\mathit{\omega }$is the angular velocity, and it is equal to the linear velocity divided by the radius of the path $\mathrm{J}/\mathrm{r}$.

$$\begin{gathered} L=\left(\frac{2}{5}m{r}^2\right)\left(\frac{v}{r}\right)=\frac{2}{5}mrg \\ \therefore r=r_e=\frac{e^2}{4\pi {\displaystyle \stackrel{‵}{o_{0}m{c}^2}}}andL=\sout{h}/2 \\ \frac{\sout{\mathrm{h}}}{2}=\frac{2}{5}\mathrm{m}\frac{{\mathrm{e}}^2}{4\mathrm{\pi }{\displaystyle \stackrel{‵}{{\mathrm{o}}_0{\mathrm{mc}}^2}}}\mathrm{\theta } \\ \sout{\mathrm{h}}=\frac{{\mathrm{e}}^{-\mathrm{t}}}{5\mathrm{\pi }\stackrel{‵}{{\mathrm{o}}_0{\mathrm{C}}^2}} \end{gathered}$$

Thus, the expression of Plank’s constant is obtained.

In this step substitute the given values in the expression of velocity,

$$\begin{gathered} \mathrm{\vartheta }=\frac{5\mathrm{\pi h}{\displaystyle \stackrel{\text{'}}{\mathrm{o}}}{\mathrm{c}}^2}{{\mathrm{e}}^2} \\ =\frac{5\mathrm{\pi }\left(1.055\times {10}^{-34}\right)\left(8.85\times {10}^{-12}\right){\left(3\times {10}^8\right)}^2}{\left(1.6\times {10}^{-19}\right)2} \\ =5.156\times {10}^{10}\mathrm{m}/\mathrm{s}. \end{gathered}$$

If the quantum spin were the same as the classical spin, one of the postulates of special relativity would be violated (i.e.,$\mathrm{\vartheta }>\mathrm{C}$), which couldn't be true because nothing could move faster than the speed of light, hence the quantum spin must be something altogether different.

As nothing can travel faster than the speed of light, quantum spin is distinct from classical spin.