# Nuclear Fusion Reaction [closed]

My question is about the energy released in a nuclear fusion reaction, consider:

$\mathrm{{^{2}_1H}} \ + \ \mathrm{{^{2}_1H}} \ \rightarrow \ \mathrm{{^{3}_2He} \ + \ n \ + \ 3.27 \ MeV}$

This page shows that this energy is released as the kinetic energy of the products of the reaction. I was wondering from where the heat and light energy is coming in sun through these reactions, if whole of the energy released is carried as the kinetic energy of the products? Explain.

## closed as off-topic by AccidentalFourierTransform, Bill N, Jon Custer, Daniel Griscom, glSJan 12 '18 at 9:38

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It has a specific spectrum and intensity that depends only on the body's temperature, which is assumed for the sake of calculations and theory to be uniform and constant.

It has been observed that all bodies radiate electromagnetic radiation according to the formula given in Gijsv's answer.

It is because all matter is made up of atoms and atoms have charges and magnetic and electric distributions . Atoms hitting each other do not just bounce, exchanging energy , but also, due to the electric and magnetic fields also radiate away some of the energy until they reach an equilibrium.

In the core of the sun, the atoms are ionized, it is a soup of electrons and ions, so the electric fields are strong . The helium and neutrons(neutron has a magnetic moment) of the fusion reaction hit around and lose their kinetic energy in interacting with the electric and magnetic fields of each other, emitting photons, until they come into equilibrium with the surrounding soup .

The sun energy production is a complicated story, but the rough description holds.

The alpha particles that are produced have a high kinetic energy, this high kinetic energy will translate to a high temperature, in fact the sun has a very high temperature in the core of 10 to 15 million kelvin. $^1$ These very hot bodies will emit black body radiation which is light and do not directly produce photons as for example an laser. The sun is basically a very hot light bulb. The wavelength and intensity that a black body emits is determined by it's temperature according to the Planck's law: $$B_\nu(\nu, T) = \frac{ 2 h \nu^3}{c^2} \frac{1}{e^\frac{h\nu}{k_\mathrm B T} - 1}.$$ Further reading

in addition to the kinetic energy of the fusion reaction products created in the core of the sun, electromagnetic energy is also released in the form of gamma ray photons which propagate out from the core. they then experience nearly countless collisions with nuclei on their way out. Many of those collisions are inelastic in the sense that energy is imparted to the nuclei and subtracted from the photons, so that by the time the photons make it out to the edge of the photosphere and propagate away into space, their per-photon energy has been reduced and distributed into the blackbody spectrum we observe from here.