Question

The electron is known to have a radius no larger than ${\mathbf{10}}^{\mathbf{-}\mathbf{18}}\mathbf{\text{m}}$. If actually produced by circulating mass, its intrinsic angular momentum of roughlywould imply very high speed, even if all that mass were as far from the axis as possible.

(a) Using simply$\mathbf{rp}$(from |r × p|) for the angular momentum of a mass at radius r, obtain a rough value of p and show that it would imply a highly relativistic speed.

(b) At such speeds,$\mathbf{E}\mathbf{=}{\mathbf{\gamma mc}}^{\mathbf{2}}$and$\mathbf{p}\mathbf{=}\mathbf{\gamma mu}$combine to give$\mathbf{E}\mathbf{\cong }\mathbf{pc}$(just as for the speedy photon). How does this energy compare with the known internal energy of the electron?

Short answer

1. The rough value of p is$1.055\times {10}^{-16}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$.
2. This energy is much greater than the known internal energy of the electron.

Step-by-step solution

Definition of spin angular momentum

The spin angular momentum of light (SAM) is the component of the angular momentum of light that is associated with the quantum spin and the rotation between the polarization degrees of freedom of the photon.

Aim of the question

This question aims to prove that the intrinsic/spin angular momentum$\left(\mathrm{S}\right)$ of the electron does not have a classical analog. Part (a) of this question instructs us to use the simple formula for angular momentum $(\mathrm{L}=\mathrm{r}\times \mathrm{p})$to prove that such an angular momentum of a value roughly equaling$\mathrm{h}$ would imply a relativistic speed. Part (b) asks us to compare the relativistic energy with the known internal/rest energy of the electron.

Determine the value of p(a)

Taking the intrinsic angular momentum of the electron $(\mathrm{S}\simeq \mathrm{h})$to simply equal $\mathrm{rp}$where the electron's radius is taken to be ${10}^{-18}\mathrm{m}$, we can calculate the momentum $\mathrm{p}$as follows,

$\mathrm{S}=\mathrm{rp}$

$\mathrm{\hslash }=\left({10}^{-18}\mathrm{m}\right)\mathrm{p}$

$\begin{array}{c}\mathrm{p}=\frac{\mathrm{\hslash }}{{10}^{-18}\mathrm{m}}\\ =\frac{1.005\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}}{{10}^{-18}\mathrm{m}}\\ =1.055\times {10}^{-16}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}\mathrm{.............................}\left(2\right)\end{array}$

If we treat$\mathrm{p}$it classically, this would mean that the electron's speed has the value

$\begin{array}{c}\mathrm{u}=\frac{\mathrm{p}}{\mathrm{m}}\\ =\frac{1.055\times {10}^{-16}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}}{9.1\times {10}^{-31}\mathrm{kg}}\\ =1.16\times {10}^{14}\mathrm{m}/\mathrm{s}\end{array}$

This value is much greater than the speed of light $(\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s})$which is not allowed according to the special theory of relativity. So next, we try to treat $p$relativistically to get $\mathrm{u}$.

The relativistic momentum is given by $\mathrm{p}=\mathrm{\gamma mu}$where $\mathrm{\gamma }=\sqrt{\frac{1}{1-{\mathrm{u}}^2/{\mathrm{c}}^2}}$.

Accordingly, solving for $\mathrm{u}$ gives us the following formula,

$\mathrm{u}=\sqrt{\frac{{(\mathrm{p}/\mathrm{m})}^2}{1+{(\mathrm{p}/\mathrm{m})}^2/{\mathrm{c}}^2}}$

We have already calculated$\mathrm{p}/{\mathrm{m}}_{\mathrm{e}}$ in our failed attempt to treat $\mathrm{p}$classically. Substituting then- gives us

$\begin{array}{c}\mathrm{u}=\sqrt{\frac{{(1.16\times {10}^{14}\mathrm{m}/\mathrm{s})}^2}{1+{(1.16\times {10}^{14}\mathrm{m}/\mathrm{s})}^2/{(3\times {10}^8\mathrm{m}/\mathrm{s})}^2}}\\ =3\times {10}^8\mathrm{m}/\mathrm{s}\end{array}$

Which equals the speed of light!

Therefore the rough value of p is$1.055\times {10}^{-16}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$.

Energy Comparison(b)

At such high relativistic speeds, the electron's energy can be approximated as

$\begin{array}{c}\mathrm{E}\cong \mathrm{pc}\\ =(1.055\times {10}^{-16}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s})(3\times {10}^8\mathrm{m}/\mathrm{s})\\ =3.165\times {10}^{-8}\mathrm{J}\\ =2\times {10}^{11}\mathrm{eV}\end{array}$

Whereas the rest energy of the electron is

$\begin{array}{c}{\mathrm{E}}_{\mathrm{o}}={\mathrm{mc}}^2\\ =(9.1\times {10}^{-31}\mathrm{kg}){(3\times {10}^8\mathrm{m}/\mathrm{s})}^2\\ =8.19\times {10}^{-14}\mathrm{J}\\ =5.1\times {10}^5\mathrm{eV}\end{array}$

Accordingly,$\mathrm{E}>>{\mathrm{E}}_{\mathrm{o}}$ and the approximation$\mathrm{E}\cong \mathrm{pc}$ is justified.

These results imply that the electron behaves similar to a with speed which is not the reality of the situation

Therefore energy is much greater than the known internal energy of the electron.