Question #2b0c9

So, you know that the first nuclear equation has aluminium-27 react with a neutron.

A neutron has no charge and an atomic mass equal to one, so you can write

#""_13^27"Al" + ""_0^1"n" -> ""_13^28"Al"#

The resulting aluminium-28 isotope will then decay via alpha decay.

When isotopes undergo alpah decay, their nucleus emits an alpha particle, which is essentially the nucleus of a helium-4 atom.

Helium-4 has an atomic number equal to #2# and a mass number equal to #4#. This means that you can write

#""_13^28"Al" -> ""_Z^A"X" + ""_2^4"He"#

Since atomic number and mass number must be conserved in nuclear equations, you get that element #"X"# will have

#A = 28 - 4 = 24 " "# and #" "Z = 13 - 2 = 11#

The element that has an atomic number equal to #11# is sodium, #"Na"#, which means that you are dealing with the sodium-24 isotope

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

Finally, the sodium-24 isotope undergoes beta decay. In sodium-24's case, it's beta minus, #beta^(-)# decay.

During beta decay, a neutron from the nucleus of the radioactive isotope is converted into a proton. In addition to this, an electron and an electron antineutrino, #bar(nu)_e#, are emitted by the nucleus.

So, if a neutron is converted into a proton, you can expect the atomic number to increase by #1#, but the mass number to remain unchanged.

This means that you have

#""_11^24"Na" -> ""_12^24"Y" + ""_text(-1)^0"e" + bar(nu)_e#

The element that has 12 protons in its nucleus is magnesium, which means that the final product of this decay chain is the magnesium-24 isotope.

#""_11^24"Na" -> ""_12^24"Mg" + ""_text(-1)^0"e" + bar(nu)_e#