Discover the Velocities of Nucleons | Proton, Neutron and More

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SUMMARY

The velocities of nucleons, specifically protons and neutrons in Deuterium (Hydrogen-2), can be derived from their kinetic energy and binding energy. For the deuteron, the total kinetic energy is calculated as 35.7 MeV, based on a binding energy of 2.22 MeV and a potential well depth of 38.5 MeV. For arbitrary nuclei, the nuclear radius is approximated by r = r0A1/3, where r0 is 1.25 fermi. The Fermi momentum, kF, is determined to be 1.6 fm-1, leading to a momentum of 316 MeV, which is one third of the rest energy of a proton or neutron.

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jaketodd
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Thanks everyone for your help over the years...

Much appreciated if someone will tell me:

A) The velocities of the proton and neutron in Deuterium/Hydrogen2

B) How to calculate the velocities of nucleons for arbitrary atoms
 
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Do you mean how fast they're moving around...? Is that a meaningful concept for nucleons? They are confined in a very small volume, and by the uncertainty principle, the uncertainty in their momentum/velocity will be very large.
 
For the deuteron it's easy to find the experimental numbers. It's a triplet S state with binding energy 2.22 MeV. The best fit for the potential well has depth 38.5 MeV, meaning the total kinetic energy is 38.5 - 2.22 = 35.7 MeV. (Do you really want the velocity, or is that good enough?)

For an arbitrary nucleus it's harder to make a good estimate. The nuclear radius is roughly r = r0A1/3 where r0 = 1.25 f. The potential is somewhere between harmonic oscillator and square well with a depth typically 50 MeV. If you want the total KE or the average KE don't just take the lowest level in this well, remember the nucleons will occupy the well states up to some highest level.
 
Assume the nuclear matter (protons and neutrons as indistinguishable) forms a degenerate fermi gas. The fermi momentum is determined by:

<br /> A = 2 \, \frac{4 \pi R^{3}}{3} \, \frac{4 \pi k^{3}_{F}} {3} \, \frac{1}{(2\pi)^{3}} = \frac{4}{9<br /> \pi} (k_{F} R)^{3} \Rightarrow R = \left(\frac{9\pi}{4}\right)^{\frac{1}{3}} \frac{A^{1/3}}{k_{F}} = R_{0} \, A^{1/3}
<br /> R_{0} = \left(\frac{9\pi}{4}\right)^{\frac{1}{3}} \frac{1}{k_{F}} \Rightarrow k_{F} = \left(\frac{9\pi}{4}\right)^{\frac{1}{3}} R^{-1}_{0}<br />
Using the emprical result R_{0} = 1.2 \, \mathrm{fm}, we get:
<br /> k_{F} = 1.6 \, \mathrm{fm}^{-1}<br />
The momentum corresponding to this is:
<br /> p_{F} \, c = \hbar \, c \, k_{F} = 316 \, \mathrm{MeV}<br />
which is one third of the rest energy of a proton (neutron). That is why one should use relativistic equation:
<br /> p = m \, c \, \beta, \gamma, \ \gamma = (1 - \beta)^{-1/2}, \; \beta = v/c<br />
Then, use the fact that you have a FD distribution in momentum space to find the distribution in velocity space. From this distribution you can find the most probable speed, the average speed and the root mean square speed, for example. A rough estimate, however is to simply use the fermi momentum:
<br /> \beta_{F} = v_{F}/ c = \frac{p_{F}/(m \, c)}{\sqrt{1 + (p_{F}/(m \, c))^{2}}} = 0.32<br />
 
Last edited:
Thanks all you guys and/or gals! =)
 

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