What nuclide undergoes alpha decay to produce sodium-24?

1 Answer
Sep 30, 2015

Answer:

#""_13^28"Al"#

Explanation:

You know that alpha decay takes place when an #alpha"-particle"# is being ejected from the nucleus of a radioactive isotope.

http://scienceblogs.com/startswithabang/2012/12/12/why-the-world-will-run-out-of-helium/

An #alpha"-particle"# is simply the nucleus of a helium-4 atom. A helium-4 atom has a total of two protons and two neutrons in its nucleus, and two electrons surrounding that nucleus.

In that case, if an #alpha"-particle"# is the nucleus of a helium-4 atom, then it must have a mass number equal to #4#, since it has two protons and two neutrons, and a net charge of #(2+)#, since it no longer has the two electrons that the helium-4 atom has.

So, you know that an unknown radioactive isotope decays via alpha decay to yield a sodium-24 nucleus.

A sodium-24 nucleus contains 11 protons and 13 neutrons. This means that you can write

#""_Z^A"X" -> ""_11^24"Na" + ""_2^4alpha#

Here #A# and #Z# represent the unknown element's mass number and atomic number, respectively.

So, if you take into account the fact that mass number and atomic number must be conserved in a nuclear reaction, you an say that

#A = 24 + 4" "# and #" "Z = 11 + 2#

The unknown isotope will thus have

#{(A = 28), (Z= 13) :}#

A quick look in the periodic table will show you that the element that has 13 protons in its nucleus is aluminium, #"Al"#. This means that you're dealing with aluminium-28, an isotope of aluminium that has 15 protons in its nucleus.

The complete nuclear equation will be

#""_13^28"Al" -> ""_11^24"Na" + ""_2^4alpha#