# 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?

Jun 21, 2016

${\text{_ (color(white)(a)92)^234"U" -> ""_ (color(white)(a)90)^230"Th" + }}_{2}^{4} \alpha$

#### 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, $\text{^4"He}$, 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.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} A = Z + N \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

$A = \text{92 protons" + "142 neutrons} = 234$

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, $\text{U}$, and so you have $\text{_ (color(white)(a)92)^234"U}$ 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

${\text{_ (color(white)(a)92)^234"U" -> ""_m^n"X" + }}_{2}^{4} \alpha$

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

$92 = m + 2 \text{ "->" }$ conservation of charge

$234 = n + 4 \text{ "->" }$ conservation of mass

You will thus have

$m = 92 - 2 = 90 \text{ }$ and $\text{ } n = 234 - 4 = 230$

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

The balanced nuclear equation will thus be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{_ (color(white)(a)92)^234"U" " "->" " ""_ (color(white)(a)90)^230"Th" " "+" " }}_{2}^{4} \alpha} \textcolor{w h i t e}{\frac{a}{a}} |}}}$