Question

A radioactive nucleus alpha decays to yield a sodium-24 nucleus in 14.8 hours. What was the identity of the original nucleus? Show the nuclear equation that leads you to this answer.

Answer 1

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

An α-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 α-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 chargeof (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

^A_ZX→²⁴₁₁Na+⁴₂α

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=28Z=13

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

The complete nuclear equation will be

²⁸₁₃Al→²⁴₁₁Na+⁴₂α
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