# How do you write nuclear equations for alpha decay?

Jan 22, 2018

As an example: ${\text{_86^219 Rn -> ""_84^215 Po + }}_{2}^{4} H e$

#### Explanation:

In alpha decay, two protons and two neutrons are released by a radioactive isotopes nucleus. This means we then produce a new nucleus with two less protons and two less neutrons in then the radioactive isotope before the alpha decay process occurred.

For example;

${\text{_86^219 Rn -> ""_84^215 Po + }}_{2}^{4} H e$

Here, we start with an isotope of Rn with a relative atomic mass of 219 and an atomic number of 86.

(relative atomic mass being the number of protons and neutrons in an atom, atomic number being the number of protons in an atom).

In the decay, the Rn isotope loses 2 protons and 2 neutrons from it's nucleus. Therefore we now have an atom that has a relative atomic mass of 215 and an atomic number of 84.

If you change the atomic number of an element, it becomes a different element! This is because every element has a different number of protons (atomic number). So the alpha decay process, has in this case, basically produced an isotope of Po.

We have to account for the 'alpha particle' produced (consisting of two protons and two neutrons) and as it is technically has the same relative atomic mass and atomic number as a Helium (He) atoms nucleus, we can call the particle 'He' in the equation.