Question

(a) If you treat an electron as a classical spherical object with a radius of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{17}}\mathbf{}\mathit{m}$, what angular speed is necessary to produce a spin angular momentum of magnitude $\sqrt{\frac{\mathbf{3}}{\mathbf{4}}}\mathit{ħ}$? (b) Use$\mathbf{v}\mathbf{=}\mathbf{r\omega }$ and the result of part (a) to calculate the speed v of a point at the electron’s equator. What does your result suggest about the validity of this model?

Short answer

(a) Angular speed necessary to produce spin angular momentum of$\sqrt{\frac{3}{4}}ħ$ is $2.51\times {10}^{30}\hspace{0.33em}\mathrm{rad}/\mathrm{s}$.

(b) Speed of a point at electron’s equator is $2.51\times {10}^{13}m/s$. The speed is greater than speed of light which can never be possible as per special theory of relativity so this model is invalid.

Step-by-step solution

Physical quantities related to classical spherical object

Moment of inertia of spherical body ( $\mathbf{radius}\mathbf{=}\mathbf{R}\mathbf{,}\mathbf{\hspace{0.33em}}\mathbf{mass}\mathbf{=}\mathbf{M}$) rotating around a symmetrical axis is,

$\mathbf{l}\mathbf{=}\frac{\mathbf{2}}{\mathbf{5}}{\mathbf{MR}}^{\mathbf{2}}$

Angular momentum of rigid body with moment of inertiarotating around a symmetrical axis with angular velocity$\mathbf{\omega }$ is,

$\mathbf{L}\mathbf{=}\mathbf{l\omega }$

Calculation

Treat an electron as a classical spherical object with a radius of$\mathbf{R}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{17}}\mathbf{}\mathbf{\hspace{0.33em}}\mathbf{m}$ and$\mathbf{M}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{11}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{\hspace{0.33em}}\mathbf{kg}$

The moment of inertia is,

$$\begin{gathered} \mathrm{l}=\frac{2}{5}{\mathrm{MR}}^2 \\ \mathrm{l}=\frac{2}{5}\left(9.11\times {10}^{-31}\mathrm{kg}\right)\times {\left(1.0\times {10}^{-17}\mathrm{m}\right)}^2 \\ \mathrm{l}=3.64\times {10}^{-65}\mathrm{kg}.{\mathrm{m}}^2 \end{gathered}$$

To produce a spin angular momentum of magnitude $L=\sqrt{\frac{3}{4}}ħ$, necessary angular velocity$\omega$ is,

$$\begin{gathered} \omega =\frac{L}{l} \\ \omega =\frac{\sqrt{{\displaystyle \frac{3}{4}}}\times \left(1.055\times {10}^{-34}J.s\right)}{3.64\times {10}^{-65}kg.m^2} \\ \omega =2.51\times {10}^{30}rad/s \end{gathered}$$

Using the relation,

$\mathrm{v}=\mathrm{r\omega }$.

The speed of a point at electron’s equator is,

$$\begin{gathered} v=(1.0\times {10}^{-17}\hspace{0.33em}m)\times (2.51\times {10}^{30}\hspace{0.33em}rad/s) \\ v=2.51\times {10}^{13}\hspace{0.33em}m/s \end{gathered}$$

This model is invalid as velocity is coming to be very larger than speed of light$\mathbf{c}\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$ which is ultimate limit of speed according to postulates of special theory of relativity.

Therefore,

(a) Angular speed necessary to produce spin angular momentum of$\sqrt{\frac{3}{4}}ħ$ is$2.51\times {10}^{30}\hspace{0.33em}rad/s$

(b) Speed of a point at electron’s equator is $2.51\times {10}^{13}\hspace{0.33em}\mathrm{m}/\mathrm{s}$. The speed is greater than speed of light which can never be possible as per special theory of relativity so this model is invalid.