Question

Question #2b0c9

Answer 1

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

`""_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`