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So, during nuclear fusion, hydrogen-1 nuclei fuse with hydrogen-2 nuclei. Given that these nuclei are both positively charged, that means that they repel, and thus have a positive Coulomb Potential.($E = \frac{e^2}{4\pi\epsilon r}$)

In order for nuclear fusion to occur, high temperatures are a necessity so that each nucleus has sufficient energy to overcome the repulsive Coulomb potential.

My question is: Can we consider this kinetic energy to be 'negative'? The reason why I ask is because if we consider the kinetic energy to be positive, and the energy of the particle is the sum of the kinetic energy and the coulomb potential energy, then a rise in kinetic energy would lead to a more positive result, thus implying that it is even less bound to the other nucleus than before.

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No the kinetic energy is positive.

If we take the two nuclei as starting a large distance apart then the potential energy is effectively zero as the $r^{-2}$ term is so small. In that case the energy of the nuclei is just the sum of the kinetic energies:

$$ E = \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2 $$

As the nuclei approach each other you are quite correct that the potential energy becomes positive so the total energy is now:

$$ E = \tfrac{1}{2}m_1 {v'}_1^2 + \tfrac{1}{2}m_2 {v'}_2^2 + \frac{k Q_1 Q_2}{r_{12}^2}$$

Since energy is conserved the total energy must be the same in both cases, and for that to happen the velocities of the two nuclei must decrease i.e. $v_1' \lt v_1$ and $v_2' \lt v_2$. Which is obvious really. As the initially fast moving nuclei approach each other the repulsion between them slows them down. The increase in the positive potential energy is counterbalanced by a decrease in the positive kinetic energy.

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  • $\begingroup$ That's right but eventually it only has to get close enough to where the nuclear force takes over from the electrical force. Not sure if some tunneling might also contribute. $\endgroup$ – Bob Bee Aug 1 '17 at 15:20

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