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Qmechanic
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Colomb Coulomb barrier in nuclear fushionfusion

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 coulombCoulomb 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

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

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

Colomb barrier in nuclear fushion

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 $$2(Z-2)$$?

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 $$2(Z-2)~?$$

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Physkid
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Colomb barrier in nuclear fushion

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 $$2(Z-2)$$?