minimum energy?

Question

What's the minimum energy of a proton or neutron (assume mc^2=940MeV) confined to a region of space about the same size as a nucleus (1E-14m)?

Answer 1

Such proton has de Broglie wavelenght
λ = h/p ~ a,
and consequently momentum
p = h/λ ~ h/a.

Kinetic energy is
K = p²/(2m) ~ 1/(2m) (h/a)².

Since you have mass in units of c,
Eo = mc² = 940MeV
then
K ~ c²/(2Eo) (h/a)²
K ~ 1/(2Eo) (hc/a)².


Planck constant in units of eV is
H ~ 4e-15 eV s

K ~ 1/(2 x 940MeV) (4e-15 eV s 3e8 m/s / 1e-14 m)²

Answer:
K ~ 1/(2Eo) (hc/a)² ~ 7 MeV

Answer 2

I think Alex is correct...as far as he goes. But I'm not clear all the energy has been included in the audit. Total energy (TE) = KE + PE; where kinetic energy and potential energy are included in the audit.

Alex found KE (very nicely I might add) and PE(m) = mc^2, the potential energy of each proton/neutron mass, was given. But I'm not sure PE(r), the potential energy of a given neutron/proton due to its location r distance from the center of the nucleus was considered.

Protons and neutrons are subject to strong and weak atomic forces, which hold the core together despite the opposing positive charges of the protons. As we move out from the center (r = 0) to the surface of the nucleus (r = R = 10^-14 m, as given), more and more protons are included in the subtended volume. Also, as we move outward, the number of protons between the subtended volume and the surface of the nucleus decreases. In which case the net repulsion forces on each layer of protons increases with increasing r measured from the center.

So, my guess, PE(r) ~ np ~ rho 4/3 pi r^3 p; where n is the number of protons of charge p included in the volume of radius r (where r = R is the surface of the core), rho is the proton density (N/V; where N is the number of protons in a nucleus of V = 4/3 pi R^3 volume, where you gave R = 10^-14 m). If you know N, you can calculate rho.

That means PE(R) > PE(r < R) so that the strong and weak atomic forces (F) give each proton at R more potential energy than each proton at r < R. By example, if the protons were evenly distributed around a proton at the center (r = 0), there would be no net atomic force acting on it; so PE(r = 0) = 0 in that special case.

Bottom line, the minimum energy proton or neutron would be where TE = KE + PE(m) + PE(r = 0); where the minimum location potential energy PE(r = 0) is near or at the center of the nucleus. This is an intuitively pleasing result because we have a similar result with location potential energy PE(h) = mgh derived from gravity force. As KE and PE(m) remain the same no matter where the neutron or proton is located, TE minimum is where PE(r) is minimum (e.g., at r = 0).

One final comment, this applies to the neutrons as well, because they are also affected by the atomic strong and weak forces even though the neutrons are not the sources of the repulsion forces.

Anyway, just a thought.