How do you complete the following alpha decay reaction?

1 Answer
Jul 2, 2016

Answer:

#""_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#