# Why during nuclear fusion does the process only release neutrons? [closed]

Say we envision subatomic particles like balloons held together by static electricity (analogous to nuclear force).

Now we smash em' together. They make a new element. My chemistry test answer key has us calculate the number of neutrons that leave the reaction by using Einstein's famous equation. But how do we know they were all neutrons that left? Couldn't some protons have been bounced off into the hinterlands, taking with them the energy their mass represents?

## closed as unclear what you're asking by Loong♦, Todd Minehardt, Ben Norris, Klaus-Dieter Warzecha, F'xAug 28 '16 at 9:04

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• Looks you're talking about fusion not fission. Also it's rather physics then chemistry. – Mithoron Aug 26 '16 at 18:43
• I edited the title but you should respond. – Mithoron Aug 28 '16 at 0:46

In the chart of nuclides, the most stable nuclides have a certain ratio of proton $(Z)$ to neutron $(N)$ numbers. They form the so-called valley of β-stability. Nuclides on right side of the valley (higher neutron numbers) are unstable to decay by β emission. Nuclides on the left side are unstable to β+ decay or electron capture.

Image taken from Choppin, Liljenzin, Rydberg: Radiochemistry and Nuclear Chemistry, third edition (2002)

For the light elements, stability is achieved when the number of neutrons and protons are approximately equal $(N=Z)$, for example $\ce{^2_4He}$ with $Z=2$ and $N=2$, or $\ce{^{12}_6C}$ with $Z=6$ and $N=6$. However, with increasing atomic number of the element, the ratio of neutrons to protons for nuclear stability increases to approximately $N/Z=1.5$ for $\ce{^{209}_{83}Bi}$ with $Z=83$ and $N=126$.

The neutron excess of heavy elements explains the release of neutrons during nuclear fission as well as the radioactivity (mostly β decay) of the fission products. During the usual nuclear fission, a heavy elements is split into two smaller nuclides with a relatively large excess of neutrons. If no neutrons are released during this process, the large neutron excess of the original heavy nuclide $(N/Z\gt1.5)$ would be too large for the two smaller fission product nuclides to be stable. Hence, some of the excess neutrons are immediately released during the fission process (prompt neutrons). Further neutrons can be released later by the fission products (delayed neutrons) or are mostly converted during the beta decay of the fission products.

For example, during the nuclear fission of $\ce{^{235}_{92}U}$ with thermal neutrons, a typical fission reaction can lead to the creation of the fission products $\ce{^{140}_{54}Xe}$ and $\ce{^{93}_{38}Sr}$ and the release of three neutrons:

$$\ce{^{235}_{92}U + 1n -> ^{236}_{92}U -> ^{140}_{54}Xe + ^{93}_{38}Sr + 3n}$$

The neutron excess of $\ce{^{235}_{92}U}$ with $N=143$ and $Z=92$ is very large $(N/Z=1.54)$. It is further increased by the thermal neutron. Although three neutrons are immediately released during the fission process, the remaining neutron excess is still too large for $\ce{^{140}_{54}Xe}$ with $(N/Z=1.59)$ and $\ce{^{93}_{38}Sr}$ with $(N/Z=1.45)$ to be stable. Therefore, $\ce{^{140}_{54}Xe}$ further decays by successive β emissions, finally to the stable nuclide $\ce{^{140}_{58}Ce}$ with $(N/Z=1.41)$, and $\ce{^{93}_{38}Sr}$ finally decays to $\ce{^{93}_{41}Nb}$ with $(N/Z=1.27)$.

• This answer has become invalid when the question was changed from "fission" to "fusion". – Loong Aug 28 '16 at 1:00