Prove that p-p can't exisit

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Prove that diproton can't exisit

Homework Statement


As stated in the title I'm trying to prove that that di-proton(di-hydrogen) can't exist.
M(p)=1.008665u
M(n)=1.007825u
M(3He)=3.016030u
n-p=2.226Mev
R=R(0)*A^1/3
R(0)=1.2*10^-15

Homework Equations


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The Attempt at a Solution


What I did is that i calculated the Total binding energy of (3)He.
Bt(He) = 7.716649Mev
Bt(He) = (n-p) + (n-p) + (p-p)
so p-p= 3.2646Mev
Then I have calculated the electrostatic repulsion(U) between the protons.
these 2 protons exist at distance R=1.511*10^-15
U=0.95Mev
I've done something wrong because clearly Bt>U
Where is my mistake?
 
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Answers and Replies

  • #2
Dick
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You've only calculated the electrostatic repulsion energy for 1 proton. There are two of them acting on each other. Double your U.
 
  • #3
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the double would be U=1.9Mev , is still less then 3.2646Mev
 
  • #4
Meir Achuz
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That formula for R does not work for only two nucleons.
Your formula for Bt(He) is also oversimplified.
The simplest way to prove that a diproton doesn't exist is to use the fact that a dineutron does not exist, and a diproton would be less likely to exist because of the Coulomb repulsion.
 
  • #5
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How can I approximate the distance between the 2 protons?
 
  • #6
Dick
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This can't intended to be an exact calculation. Everything is (and has to be) oversimplified. I would think binding energy of deuterium would be a better guide for nuclear binding of a diproton than something pulled out of He(3). Hey, Meir, what would keep dineutron from beta decay? I would actually think a dineutron would be stable nuclear bindingwise. Anyway, as I recall, the diproton ALMOST exists (if nuclear binding were just a little stronger it would). So I'd be happy with just getting nuclear binding nearly the same as EM repulsion.
 
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  • #7
Dick
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There is also the rule that nuclei with an equal number of protons and neutrons are more stable than otherwise. So I would expect a diproton to be more weakly bound in terms of nuclear binding than a deutron.
 
  • #8
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You maybe on to something, The difference in Nuclear binding energy between deuterium and diproton should be in approximation very close to the difference of mass between the neutron and the proton.
That give us an amount of energy of 0.78Mev
if the n-p=2.226Mev
then I guess for diproton Bt=1.44Mev, if my calculation for the electrostatic repulsion is right U=1.9Mev, then I guess this work is ok??
 
  • #9
Dick
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You maybe on to something, The difference in Nuclear binding energy between deuterium and diproton should be in approximation very close to the difference of mass between the neutron and the proton.
That give us an amount of energy of 0.78Mev
if the n-p=2.226Mev
then I guess for diproton Bt=1.44Mev, if my calculation for the electrostatic repulsion is right U=1.9Mev, then I guess this work is ok??
I don't think the mass difference has much to do with it. That comes from quark masses and EM. I can't think of any really good way to guess the diproton nuclear binding, except that it should be 'somewhat LESS (sorry) than deuteron'. Anybody else?
 
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  • #10
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I don't think the mass difference has much to do with it. That comes from quark masses and EM. I can't think of any really good way to guess the diproton nuclear binding, except that it should be 'somewhat greater than deuteron'. Anybody else?
There maybe a way , but require some good approximations, consider that these 2 protons exist in potential that looks like the following
v(r)= -V(0) for r<r0
0 for r>r0
(taking the approximation of a finite square potential)
Solving the Wave equation , should provide us with a good approximation of the binding energy...anyone agree with this?
 
  • #11
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Calculations have led me to this:
Tg(Kr0)=-SQRT((V0-B)/B) while K is the K in c*Sin(Kr) in the solution for the wave function for r<r0, and B=-E where is E should be the binding energy
V0 I got it from a Book it should be 18Mev for Proton-Proton interaction..
and r0=2.5F
K=2*Pi*Sqrt[M(V0+E)]/h
M=2*M(Proton)
All i need to know now is Kr0 value, I see I have a very intresting equation , anyone agree with what I have done so far?
 
  • #12
Dick
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I'm pretty skeptical as to whether this can give you the sort of precision you want - it looks like more of an order of magnitude calculation. But if you are having fun, carry on.
 
  • #13
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Ok , my calculation gave me , a value of 1.078125 Mev for di-proton..
 
  • #14
Dick
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What would it give you for the deutron?
 
  • #15
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v0=36Mev , r0=2F
it equals=2.23 Mev very close to real value..
 
  • #16
Dick
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Hmm. Ok. Like I say, I thought the diproton was 'almost bound'. But then it still may be - the binding energy is probably hugely sensitive to the exact value of the strong coupling constant.
 
  • #17
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true, if the strong force was 2% higher, nuclear force would overcome the repulsion in diproton..
I still need a way to approximate the distance between the 2 protons , to calculate the repulsion between them..
 
  • #18
Meir Achuz
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The deuteron is a poor example, because it is spin one and a diproton must have spin zero. That is why the dineutron is not bound. It doesn't exist.
It's hard to estimate a distance for something that doesn't exist. The Coulomb repulsion is not the main reason that the diproton doesn't exist.
 
  • #19
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So your suggestion is to abandon the idea?
 
  • #20
Meir Achuz
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Post # 4 is the proof.
 
  • #21
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Why dineutron doesn't exist?
Now I'm very curious to know.
your talking about spin zero, care to elaborate on that a bit?what does it have to do with the fact that dineutron and diproton doesn't exist..
I know that pair neutron or protons, tend to align their spins in the opposite direction, but again what does it have to do with this?
 
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  • #22
Meir Achuz
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You have to read an intermediate book on nuclear physics.
P and n each have spin 1/2. They can combine to total spin 0 or 1.
Experimentally the deuteron is spin 1 and there is no spin 0 n-p bound state.
Because of the Pauli principle, p-p or n-n could only bind in the spin zero state. There is no dineutron, which is consistent with there being no spin zero n-p bound state. Then read post #4.
You can't use the deuteron as a starting point because the nuclear force is different for the spin zero (weaker) and the spin one cases.
 
  • #23
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I think I understand, I'm taking nuclear physics this semester, and not quiet finished understanding it entirely..
Thank you
 

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