# Coulomb barrier in nuclear fusion

Suppose there exists an alpha particle in the nucleus. Within a radius of 2 femtometer, the dominating force is the nuclear force but beyond this radius, the Coulomb force becomes the effective dominating force due to it's longer range force. Outside this radius is a daughter nucleus.

So let's suppose the alpha particle moves beyond the effective range of the nuclear force,

the electric potential energy due to the force between the alpha particle and the daughter nucleus is given by as

$$U_{B}=\frac{1}{4\pi \varepsilon _{0}}(\frac{2(Z-2)e^{2}}{R}).$$

Why is there the term $$2(Z-2)~?$$

• $Z$ is the count of protons in the parent. So the two daughter components have charges of $2e$ (the alpha) and $(Z-2)e$ (the rest). May 28, 2015 at 6:24
• @BOwlOfRed That should be an answer, not a comment.
– rob
May 28, 2015 at 13:57

The charge in the nucleus is due to protons, and the count of protons is given as $Z$ for a total charge of $Ze$.
The alpha decay removes two protons (and two uncharged neutrons). The charge of the alpha particle is $2e$, leaving a charge of $Ze - 2e$ in the nucleus after decay. Multiplying the two and rearranging, that becomes $2(Z-2)e^2$.