The mean momentum p of a nucleon in a nucleus of the mass number A and atomic number Z depends on A as P∝A^(−1/3).

Jairo Decker

Jairo Decker

Answered question

2022-10-16

The mean momentum p of a nucleon in a nucleus of the mass number A and atomic number Z depends on A as
P A 1 3 .
All that I know is that the radius of the nucleus is proportional to mass number as
R A 1 3 .
From here how can one relate momentum to the mass number?

Answer & Explanation

canhaulatlt

canhaulatlt

Beginner2022-10-17Added 17 answers

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 σ x R, and therefore momentum uncertainty
σ p / R
You can mess around with the factors of two if you like, but there's your one-squiggle proportionality relationship.
Chloe Arnold

Chloe Arnold

Beginner2022-10-18Added 6 answers

You can get a very rough idea of the momentum scales that you would expect in the nucleus simply by using the Uncertainty Principle. Essentially, when you localize a particle such that its uncertainty in position is Δ x, then its uncertainty in momentum Δ p is given by:
Δ x Δ p 2
(We use rather than 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 Δ x, we use the nuclear radius, which is roughly proportional to A 1 / 3 . So, rearranging with Δ x A 1 / 3 ,, we have that
Δ p 1 A 1 / 3 = A 1 / 3

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