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Velocities of Nucleons

by jaketodd
Tags: nucleons, velocities
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jaketodd
#1
Jun9-11, 01:15 AM
PF Gold
P: 298
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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JesseC
#2
Jun9-11, 08:20 AM
P: 281
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.
Bill_K
#3
Jun9-11, 11:18 AM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
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.

Dickfore
#4
Jun9-11, 11:40 AM
P: 3,014
Velocities of Nucleons

Assume the nuclear matter (protons and neutrons as indistinguishable) forms a degenerate fermi gas. The fermi momentum is determined by:

[tex]
A = 2 \, \frac{4 \pi R^{3}}{3} \, \frac{4 \pi k^{3}_{F}} {3} \, \frac{1}{(2\pi)^{3}} = \frac{4}{9
\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}[/tex]
[tex]
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}
[/tex]
Using the emprical result [itex]R_{0} = 1.2 \, \mathrm{fm}[/itex], we get:
[tex]
k_{F} = 1.6 \, \mathrm{fm}^{-1}
[/tex]
The momentum corresponding to this is:
[tex]
p_{F} \, c = \hbar \, c \, k_{F} = 316 \, \mathrm{MeV}
[/tex]
which is one third of the rest energy of a proton (neutron). That is why one should use relativistic equation:
[tex]
p = m \, c \, \beta, \gamma, \ \gamma = (1 - \beta)^{-1/2}, \; \beta = v/c
[/tex]
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:
[tex]
\beta_{F} = v_{F}/ c = \frac{p_{F}/(m \, c)}{\sqrt{1 + (p_{F}/(m \, c))^{2}}} = 0.32
[/tex]
jaketodd
#5
Jun9-11, 08:24 PM
PF Gold
P: 298
Thanks all you guys and/or gals! =)


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