# How do you complete the following alpha decay reaction?

Jul 2, 2016

${\text{_90^232 Th -> ""_88^228Ra + }}_{2}^{4} H e$

#### Explanation:

The general notation of a nuclide ($X$) is:

""_Z^AX

In wich $Z$ is the number of protons and $A$ the mass number (protons + neutrons).

In alpha decay the nucleus emits a particle that contains 2 protons and 2 neutrons, which is similar to the nucleus of helium. So a notation for the nuclide ($Y$) that is left after emitting an alpha particle (""_2^4He) is:

${\text{_(Z-2)^(A-4)Y + }}_{2}^{4} H e$

Now you can complete the equation given in the example:

${\text{_(88+2)^(228+4)X = }}_{90}^{232} X$

The last step is finding the nuclide that has 90 protons and 142 neutrons in a table of nuclides. This appears to be Thorium ($T h$).

This makes the equation complete:

${\text{_90^232 Th -> ""_88^228Ra + }}_{2}^{4} H e$