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The force required to attach two protons?

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There are two protons in the room temperature.

They have repulsion force each other.

Between them no fusion reaction occur.

If we only consider proton as particle, how much force is required for attaching them each other?

Edited by alpha2cen

They need to overcome the Coulomb barrier, i.e. the electrostatic force of repulsion, at the distance where the attractive nuclear interaction takes over. F = kQ1Q2/r^2

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The equation states r~=0 infinite force is required.

It means the attachment is almost impossible.

Does it obtain that value in actual circumstance?

Edited by alpha2cen

They need to overcome the Coulomb barrier, i.e. the electrostatic force of repulsion, at the distance where the attractive nuclear interaction takes over. F = kQ1Q2/r^2

 

Hey swan, there was some lecture I went to and in it was a theory on how fusion in the sun occurs. I didn't get a good look at the math, but they said that the force in the Sun isn't enough to force particles together and cause them to fuse, so they instead said that the reason particles such as protons fuse is because when they get into a close enough proximity to each other, their highest areas probabilities overlap each other's boundaries where the electro-magnetic force would push them away (or something similar to that). Is that at all possible?

Hey swan, there was some lecture I went to and in it was a theory on how fusion in the sun occurs. I didn't get a good look at the math, but they said that the force in the Sun isn't enough to force particles together and cause them to fuse, so they instead said that the reason particles such as protons fuse is because when they get into a close enough proximity to each other, their highest areas probabilities overlap each other's boundaries where the electro-magnetic force would push them away (or something similar to that). Is that at all possible?

 

Right. The next effect one has to consider is quantum mechanical tunneling, so fusion can occur at larger separations (and lower temperatures) than are predicted classically.

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