Question

A certain element is radioactive. It emits alpha particles. Each of its nuclei contains 92 protons and 142 neutrons. Write a nuclear equation that describes the change that takes place by alpha emission?

Answer 1

`""_ (color(white)(a)92)^234"U" -> ""_ (color(white)(a)90)^230"Th" + ""_2^4alpha`

Explanation:

The first thing to do here is make sure that you have a clear understanding of what an alpha particle is.

An alpha particles is simply the nucleus of a helium-4 atom, `""^4"He"`, and it contains `2` protons and `2` neutrons.

Now, your next goal here is to figure out the mass number of the radioactive isotope given to you by the problem.

As you know, the mass number, `A`, is calculated by adding the number of protons, i.e. the atomic number, `Z`, and the number of neutrons, `N`, present in a nucleus.

`color(blue)(|bar(ul(color(white)(a/a)A = Z + Ncolor(white)(a/a)|)))`

In your case, you have `92` protons and `142` neutrons in the nucleus, which means that the nuclide's mass number will be

`A = "92 protons" + "142 neutrons" = 234`

Next, use isotope notation to write the isotopic symbol for this element

A quick look in the periodic table will reveal that your unknown element is uranium, `"U"`, and so you have `""_ (color(white)(a)92)^234"U"` as the isotope notation here.

Now, when a nuclide emits an alpha particle, two things happen

• its atomic number decreases by `2` because an lpha particle contains `2` protons
• its mass number decreases by `4` because an alpha particle contains `2` protons and `2` neutrons

You can thus set up your nuclear equation like this

`""_ (color(white)(a)92)^234"U" -> ""_m^n"X" + ""_2^4alpha`

In any nuclear reaction, charge and mass are conserved, which means that you can write

`92 = m + 2 " "->" "` conservation of charge

`234 = n + 4 " "->" "` conservation of mass

You will thus have

`m = 92 - 2 = 90" "` and `" "n = 234 - 4 = 230`

Another quick look in the periodic table will reveal that the element that has `90` protons in its nucleus is thorium, `"Th"`. The nuclide produced by the alpha decay will thus be thorium-230, `""_ (color(white)(a)90)^230"Th"`.

The balanced nuclear equation will thus be

`color(green)(|bar(ul(color(white)(a/a)color(black)(""_ (color(white)(a)92)^234"U" " "->" " ""_ (color(white)(a)90)^230"Th" " "+" " ""_2^4alpha)color(white)(a/a)|)))`