Question

How do you complete the following alpha decay reaction?

Answer 1

`""_90^232 Th -> ""_88^228Ra + ""_2^4He`

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:

`""_(Z-2)^(A-4)Y + ""_2^4He`

Now you can complete the equation given in the example:

`""_(88+2)^(228+4)X = ""_90^232X`

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

This makes the equation complete:

`""_90^232 Th -> ""_88^228Ra + ""_2^4He`