Question:

The proton’s mass and radius are 1.67 \(\times\) 10\(^{-27}\) kg and 1.00 \(\times\) 10\(^{-15}\) m. This system is governed by quantum and classical regimes. Namely, the proton is assumed to have a rotational uniform solid spherical body and a half spin of quantum angular momentum. Find the equatorial velocity of the proton.

Answer:

In terms of classical mechanics, since it is a sphere, the moment of inertia of proton is

\[ I = \frac{2}{5}mr^2 \]

Therefore, the angular momentum becomes

\[ L = I\omega = \frac{2}{5}mr^2\omega \]

For the quantum mechanics, the spin angular momentum is given as

\[ S = \frac{1}{2}\hbar \]

Therefore, you can equate them:

\[ \frac{2}{5}mr^2\omega  = \frac{1}{2}\hbar \]

Solve for the angular velocity.

\[ \omega = \frac{1}{2}\hbar \frac{5}{2mr^2} \]

The equatorial velocity is given by

\[ v = r\omega =  \frac{5\hbar}{4mr} \]

\[     = \frac{5}{4}\frac{1.055\times 10^{-34}\mathrm{Js}}{1.673\times 10^{-27}\mathrm{kg}\times 1.00\times 10^{-15}\mathrm{m}} \]

Then we have

\[ v = 7.88 \times 10^7 \mathrm{m/s} \]

It is very fast!!!