# Matter to energy conversion

Okay so I have a question, during a nuclear explosion or particle/antiparticle annihilation, matter is converted into energy. How do I determine if a from a explosion will come lets say a small amount of high energy photons like gamma rays or a large number of low energy photons like visible light? So basically, how do I know what type of energy is the matter gonna be converted to, whether gamma rays or visible light or both?

• "matter is converted into energy" Beware of this misleading statement; it is bound to create confusion as it seems to imply new energy got created from matter and total energy of the system increased. Better to say that energy changed form from the rest energy of matter to EM energy of radiation. – Ján Lalinský Aug 23 '15 at 12:28

## 2 Answers

This is a special example of "what will happen" under given circumstances. Almost all of physics – and natural science – is about answering such questions. But they're really very many very different questions and one must be a little bit more specific about what the question is. Your general question "what forms of energy will result" is so general that it basically covers all of physics, too. At least, you must define the initial state accurately enough.

The annihilation of matter with antimatter usually produces particles of rather fixed energies whose $E$ is comparable to $mc^2$ of the annihilated particle. In particular, one electron and one positron (its antiparticle) is "very likely" to annihilate into a pair of photons. Each of the photons has the energy of 511 keV. The probability (cross section) of such annihilation may be calculated from the theory known as "quantum electrodynamics" (or more extensive theories containing QED as a limit and/or subset). The total energy carried by such photons – gamma radiation – may be measured by "calorimeters" and their individual energies by the gamma ray spectrometer, for example.

There is basically 1 good theory to predict the "most important facts" about this process of annihilation – and many apparatuses to verify that the theory works.

The annihilation of a proton and an antiproton is more messy. There are different possibilities what can arise from that and those possibilities are "comparably likely". The simplest theory to calculate the probabilities of outcomes is "quantum chromodynamics" (QCD). The photons coming from such annihilation are below 1 GeV, even more energetic gamma rays than before.

The nuclear explosion is an even dirtier process. Heavier nuclei split into pieces – which may be done in many ways – and they produce alpha, beta, and gamma radiations. The energy from the explosion is also intensely absorbed by the surrounding air and other materials so they start to burn, increase their pressure, and create very strong light and electromagnetic radiation at pretty much all frequencies including the visible spectrum. And lots of sound waves, too.

To describe or predict what sort of energy a nuclear energy releases is an interdisciplinary problem between nuclear physics, aerodynamics, thermodynamics, and engineering which has many aspects. There is no 3-sentence answer that exactly derives what sort of energy produced from a nuclear bomb is predicted and at what amounts. To verify such predictions requires a combination of engineers, commanders, soldiers, and some people who don't care about their health too much, too.

In general, therefore, clean processes involving truly elementary particles usually produce photons at specified energy which is mostly calculable, like the gamma rays in these cases. However, collisions or fission of more complicated objects like antiprotons or even nuclear bombs produce a much more diverse collection of photons at assorted wavelengths.

Lubos's answer covers very well and in a general way the different possibilities for reactants and products in the reactions of subatomic particles and nucleons.

The basic equation is

Energy of photons = energy of reactants - energy of products.

Due to the enormous energies involved it is actually unlikely that visible light will be released, most reactions will emit in the gamma range.

Here's some examples: electron + positron --> total annhilation : 511keV (from Lubos answer)

235U --> fission to daughter nuclei: typical 202.5MeV of which 7MeV gamma, the rest kinetic energy of products. https://en.wikipedia.org/wiki/Nuclear_fission

60Co --> 60Ni: Beta decay leads to an excited form of 60Ni, which then emits two different gamma rays of approx 1Mev each: https://en.wikipedia.org/wiki/Gamma_ray

Info on the energy of fusion reactions is at https://en.wikipedia.org/wiki/Nuclear_fusion

For more http://atom.kaeri.re.kr/gamrays.html will give you a chart of what reaction corresponds to what energy, so that if you observe a gamma ray of a particular energy you can find out what reaction it came from. The lowest energy on the list is 6KeV.

In contrast, a green LED emits visible photons of about 2eV!

When you see a nuclear explosion or a star, you are not observing the gamma rays but rather the thermal radiation. Most of the radiation is thermal, in the form of lower energy infra red / visible / ultraviolet photons.

Fusion occurs deep in the sun's core, and the volumetric energy release is less than that of the human body (see reference on nuclear fusion.) Most of the gamma rays produced will be absorbed on their journey to the surface of the star, thus serving to heat the mass f the star, which then radiates thermally. Hydrogen bombs actually burn deuterium, which reacts rather quicker. Although some gamma rays can penetrate several inches of lead, it must be remembered that several miles of atmosphere has much the same stopping power.

So, given that large sources of gamma rays will fry you from a distance with their thermal output, if you want to observe gamma rays, you are better off with a small source that you can approach easily, such as a small lump of radioactive metal.