Question #2b0c9

Oct 1, 2015

Here's what I got.

Explanation:

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

$\text{_13^27"Al" + ""_0^1"n" -> ""_13^28"Al}$

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

$\text{_13^28"Al" -> ""_Z^A"X" + ""_2^4"He}$

Since atomic number and mass number must be conserved in nuclear equations, you get that element $\text{X}$ will have

$A = 28 - 4 = 24 \text{ }$ and $\text{ } Z = 13 - 2 = 11$

The element that has an atomic number equal to $11$ is sodium, $\text{Na}$, which means that you are dealing with the sodium-24 isotope

$\text{_13^28"Al" -> ""_11^24"Na" + ""_2^4"He}$

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, ${\overline{\nu}}_{e}$, 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

$\text{_11^24"Na" -> ""_12^24"Y" + ""_text(-1)^0"e} + {\overline{\nu}}_{e}$

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.

$\text{_11^24"Na" -> ""_12^24"Mg" + ""_text(-1)^0"e} + {\overline{\nu}}_{e}$