### Writing Spontaneous Fission Decay Equations

Spontaneous fission (SF) happens in heavier elements, those with atomic numbers greater than 89 and mass numbers of about 230 and above. These nuclei are unstable and decay by different forms of radioactive decay. Some by beta decay, some by positron, etc. A percentage of the time, they decay by spontaneous fission.

These percentages vary widely. In the region near 230, SF can be quite rare while in the upper regions (a 254, for example), SF is usually the most common form of decay.

Here's a brief article about SF. Here's the Wikipedia article on SF.

Example #1:

--->     +     +   4 ${\text{}}_{0}^{1}\text{n}$

Note that the atomic number is the same on both sides as well as the mass number totalling up to be the same on each side.

Nothing starts the SF from outside, the decay originates from internal imbalances.

Energy is also produced when a nuclide undergoes SF, this is typically not shown in the equation.

Example #2:

--->     +     +   3 ${\text{}}_{0}^{1}\text{n}$

Example #3:

--->     +     +   4 ${\text{}}_{0}^{1}\text{n}$

How do you know what decay products are produced? A very good question!

Go here after reading the rest of the paragraph. You will be looking at a mass distribution curve for the SF of Cf-252. There are two peaks, one for the heavier fragment and one for the lighter. In the case of Cf-252, the two mass numbers plus any neutrons released will always add up to 252, the mass number of the californium.

Example #4:

${\text{}}_{100}^{256}\text{Fm}$  --->     +

Example #5:

${\text{}}_{100}^{256}\text{Fm}$  --->     +

I could not find the mass distribution curve for Fm-256 online, but I did find references to research on that topic. Here is a search if you are interested.

Example #6: An isotope of californium-252 undergoes spontaneous fission, producing cesium-135, three neutrons and one other isotope. Determine the identity of that isotope and write out the spontaneous fission rection, using full isotopic notation.

Solution:

1) Write just the left-hand side of the equation:

--->

We know that the right-hand side must add up to 98 for the atomic number and 252 for the mass number.

2) Add in the cesium-135:

--->     +

3) Determine the other isotope:

For the atomic number, we have 98 minus 55 to equal 43, so we know the element will be Tc.

For the mass number, we have 252 minus (135 + 3) to equal 114. Remember that three neutrons are also produced.

4) Write the entire reaction equation:

--->     +     +   3 ${\text{}}_{0}^{1}\text{n}$