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What is the strength of the strong force?

  1. Jun 15, 2013 #1
    Hello everybody! :shy:
    What is the strength of the strong force between, for instance, a proton and a neutron separated by a distance on the order of femtometers (1fm and 2fm and 3fm) in newton?

    Thank you in advance! ∞
     
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  3. Jun 15, 2013 #2

    Bill_K

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    Hakim, It's a bit more complicated than that... in fact, it's a LOT more complicated. In the first place, although we often call it the nuclear force, in quantum mechanics it's really the potential V(r) that's important. Force as such is not used. Secondly, we don't typically measure things in Newtons! (Or weigh particles in kilograms.) Nuclear physics has its own set of units, much more convenient on a subatomic scale.

    The attraction between a proton and neutron can be described roughly by a square well potential, however the size and depth depend on the orientation of the particle spins. If the spins are parallel ("triplet state") the well depth is roughly 36 MeV with a radius of 2 fm (fermis = femtometers). This is deep enough to have a bound state, the deuteron. On the other hand if the spins are antiparallel ("singlet state") the well depth is less, more like 18 MeV and radius 2.5 fm. In this case there is no bound state.

    But that's only the beginning. There's a spin-orbit force present, which varies depending on the relative motion of the proton and neutron. Plus a tensor force, which is noncentral. And everything I've said relates only to isolated particles. Protons and neutrons within a nucleus are affected by forces considerably different.
     
    Last edited: Jun 15, 2013
  4. Jun 15, 2013 #3
    I done an equation that calculated the force between a proton and a neutron where the distance between their G-point is 1fm to be 9998 Newtons. Is it in the triplet state or the singlet one?
     
  5. Jun 15, 2013 #4
    Calculating the Force of attraction

    Hello!
    I just read Mass defect concept from my physics textbook. It was written that the nucleus of an atom weighs less than that of the total of the mass of the nucleons. This means that the remaining mass gets converted into energy(by Einstein's mass-energy equivalence) which we call it as binding energy of nucleus. The difference in the mass is being used as energy that holds nucleons together. For example- Binding energy of deuteron is 2.22 MeV. So I suggest you to find Binding energy of nucleons rather than finding force of attraction between them.

    For more information-http://en.wikipedia.org/wiki/Nuclear_binding_energy
     
    Last edited by a moderator: Jul 7, 2013
  6. Jun 15, 2013 #5

    mfb

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    How did you calculate that? With the model of pion exchanges?

    @DevendraC: Why do you use a different font and font size? That is harder to read.
     
  7. Jun 15, 2013 #6
    @mfb: Nope, I used my own equation... And I want to know if it is correct without having to write it for some reasons... And using 2fm it gives me: 2499 Newtons.

    Thank your for your responds. :)
     
  8. Jun 15, 2013 #7

    mfb

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    Let me guess: for 3fm, you get about 1110 N?

    That does not work. For distances larger than the diameter of the nucleons the force drops way quicker than 1/r^2.
     
  9. Jun 15, 2013 #8
    No it gives me 1.111 times 10^-12 N.
     
  10. Jun 15, 2013 #9

    mfb

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    I think you have a very strange formula.
     
  11. Jun 15, 2013 #10
    When comparing this formula with gravity it gives me exactly that the strong force is 10^38 stronger than gravity.
     
  12. Jun 15, 2013 #11

    mfb

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    Please show your formula and how you got it, or stop making some claims about a formula which are impossible to check.

    Note that the strong force is not 1038 times stronger than gravity in general.
     
  13. Jun 15, 2013 #12
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