In fusion terminology, what does it imply by taking orientational average of the fusion cross section when one of the nuclei has some static deformation/orientation? How is this average taken?

Jazlyn Raymond
2022-04-12
Answered

In fusion terminology, what does it imply by taking orientational average of the fusion cross section when one of the nuclei has some static deformation/orientation? How is this average taken?

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Braxton Gallagher

Answered 2022-04-13
Author has **21** answers

It means that you're accounting for the fact that the spins of the nuclei in the fusion interaction aren't (or are, perhaps) oriented in a particular way. For instance, consider the D-T fusion reaction

${}^{2}\mathrm{H}+{}^{3}\mathrm{H}\to {}^{4}\mathrm{H}\mathrm{e}+\mathrm{n}+18\phantom{\rule{thinmathspace}{0ex}}\mathrm{M}\mathrm{e}\mathrm{V}$

The deuteron ${}^{2}\mathrm{H}$ (or d) has nuclear spin $\hslash $, while the triton ${}^{3}\mathrm{H}$ (or t) has nuclear spin $\frac{1}{2}\hslash $. But the final state, ${}^{4}\mathrm{H}\mathrm{e}+\mathrm{n}$, has spin $\frac{1}{2}\hslash $. If the deuteron and the triton happen to interact with their spins aligned, the minimum angular momentum in the system is $\frac{3}{2}\hslash $, and the final state can't occur unless there's some orbital angular momentum to it. Most of the fusion will occur from collisions where the d and t spins happen to be anti-aligned with each other. Neglecting to account the times when the d and t are aligned parallel, and the fusion coefficient is much smaller, would overestimate the cross section by a factor of $\frac{2}{3}$, or thereabouts.

${}^{2}\mathrm{H}+{}^{3}\mathrm{H}\to {}^{4}\mathrm{H}\mathrm{e}+\mathrm{n}+18\phantom{\rule{thinmathspace}{0ex}}\mathrm{M}\mathrm{e}\mathrm{V}$

The deuteron ${}^{2}\mathrm{H}$ (or d) has nuclear spin $\hslash $, while the triton ${}^{3}\mathrm{H}$ (or t) has nuclear spin $\frac{1}{2}\hslash $. But the final state, ${}^{4}\mathrm{H}\mathrm{e}+\mathrm{n}$, has spin $\frac{1}{2}\hslash $. If the deuteron and the triton happen to interact with their spins aligned, the minimum angular momentum in the system is $\frac{3}{2}\hslash $, and the final state can't occur unless there's some orbital angular momentum to it. Most of the fusion will occur from collisions where the d and t spins happen to be anti-aligned with each other. Neglecting to account the times when the d and t are aligned parallel, and the fusion coefficient is much smaller, would overestimate the cross section by a factor of $\frac{2}{3}$, or thereabouts.

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In nuclear fusion, the colliding particle unites with the parent and fuses into a single nucleus with a higher mass. Sometimes other small particles (n, p) are given off.

However, that is not the case for the reaction:

${}_{14}^{7}\text{N}{+}_{1}^{0}\text{n}{\to}_{14}^{6}\text{C}+{}_{1}^{1}\text{H}$

If so, why would you still consider it as a nuclear fusion? I have come up with a theory that it is because the carbon has an extra neutron than Nitrogen. Is this correct?

However, that is not the case for the reaction:

${}_{14}^{7}\text{N}{+}_{1}^{0}\text{n}{\to}_{14}^{6}\text{C}+{}_{1}^{1}\text{H}$

If so, why would you still consider it as a nuclear fusion? I have come up with a theory that it is because the carbon has an extra neutron than Nitrogen. Is this correct?

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Basically, how are the nuclear fusion cross sections calculated and are they a function of temperature?

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