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Calculating the Nuclear Strong Force?

by 10Exahertz
Tags: claculation, nuclear, strong force
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10Exahertz
#1
Sep12-13, 05:55 PM
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For an Ap Physics Homework assignment our teacher had us calculate the Nuclear Strong Force for a Helium nucleus with a diameter of 2 femtometers. However, the intention was to calculate the nuclear strong force by assuming that the nucleus is intact, in such a case the particles would not be changing their distance from one another. Else, one force would overpower the other and the nucleus would either fuse or separate. Therefore, one can calculate the strong force by making it equal to the electromagnetic force. I however, do not believe this is an accurate way to calculate the nuclear strong force because the nuclear strong force does not only apply to the protons in the nucleus it also applies to the neutrons. Also, inside of a nucleus, if we consider a knowledge of the position of the particles (assuming the forces are equalized and the particles not moving, relative to each other) wouldn't the Uncertainty Principle dictate that we would not know the momentum of these particles at all, and if so would this have an effect on this calculation, using this method. My overall greatest concern with answering this question with the method suggested is that a Helium Nucleus is normally 1 femtometer in diameter. So, if the Helium nucleus in this case is at 2 femtometers in diameter wouldn't the nucleus be dissociating. I know that the nuclear strong force is not like the true strong force and loses strength as the distance increases, so if the nucleus is at a greater diameter and is separating, the method used is inaccurate, right?
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mfb
#2
Sep12-13, 06:34 PM
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Therefore, one can calculate the strong force by making it equal to the electromagnetic force.
That is a completely wrong approach.

Also, inside of a nucleus, if we consider a knowledge of the position of the particles (assuming the forces are equalized and the particles not moving, relative to each other) wouldn't the Uncertainty Principle dictate that we would not know the momentum of these particles at all, and if so would this have an effect on this calculation, using this method.
The uncertainty principle is not a big issue for nucleons, they are heavy.
But those nucleons are not billard balls, they don't have fixed positions in the nucleus - they have wave-functions, distributed over the whole nucleus.

My overall greatest concern with answering this question with the method suggested is that a Helium Nucleus is normally 1 femtometer in diameter. So, if the Helium nucleus in this case is at 2 femtometers in diameter wouldn't the nucleus be dissociating.
Well, I don't think that is the main (or even an important) issue relative to the whole idea of considering the electromagnetic and the residual strong force as equal.


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