How does the mean momentum of a nucleon in a nucleus of mass number $A$ and atomic number $Z$ depends on $A$? I was going through a text and it said that

The mean momentum $p$ of a nucleon in a nucleus of the mass number $A$ and atomic
number $Z$ depends on $A$ as $$
P\propto{A}^{-\frac{1}{3}}.
$$

All that I know is that the radius of the nucleus is proportional to mass number as  $$
R \propto A^{\frac{1}{3}}.
$$ From here how can one relate momentum to the mass number? I would be grateful for even a hint.
 A: You have an answer about nuclei with angular momentum. But the result also holds for spinless nuclei, because of the uncertainty principle. A particle that is somewhere within a box of size $R$ has position uncertainty $\sigma_x \sim R$, and therefore momentum uncertainty
$$ 
\sigma_p \gtrsim \hbar/R
$$
You can mess around with the factors of two if you like, but there's your one-squiggle proportionality relationship.
A: You can get a very rough idea of the momentum scales that you would expect in the nucleus simply by using the Heisenberg Uncertainty Principle. Essentially, when you localize a particle such that its uncertainty in position is $\Delta x$, then its uncertainty in momentum $\Delta p$ is given by:
$$\Delta x\Delta p\gtrsim\frac{\hbar}{2}$$
(We use $\gtrsim$ rather than $\geq$ here because we're not quite localizing a particle to a particular 1-D box, or even to a 3-D cube; the nucleus is closer to a spherical well, and the operator structure is somewhat different. That said, it shouldn't be too different if we only want a very rough estimate. And really, all that matters for the purposes of our question is that the right-hand side is a constant.)
In the rest frame of the nucleus, the average momentum vector of a nucleon in the nucleus is zero, so the uncertainty in the momentum gives you a very rough measure of the magnitude of the momentum of the average nucleon. For $\Delta x$, we use the nuclear radius, which is roughly proportional to $A^{1/3}$. So, rearranging with $\Delta x\propto A^{1/3},$ we have that
$$\Delta p\propto \frac{1}{A^{1/3}}=A^{-1/3}$$
A: I think it's simple :
We have:
$$R \propto A^{\frac{1}3}$$
Which is exactly in my textbook:
$$R\approx1.4\times10^{-15}\sqrt[3]{A}$$
And the "angular momentum" is :
$$p=I\omega$$
where $I$ is the moment of inertia and $\displaystyle \omega=\frac{v}R$
I think it's obvious now:
$$p\propto A^{-\frac{1}3}$$
Isn't it?
