# What is the radioactive particle released in the following nuclear equation ""_74^159"W" -> ""_72^155"Hf" + ?

Jul 10, 2016

""_2^4alpha

#### Explanation:

The nuclear equation given to you describes the decay of tungsten-159 to hafnium-155

""_ (color(white)(a)74)^159"W" -> ""_ (color(white)(a)72)^155"Hf" + ""_Z^A?

Your goal here is to find the values of $A$ and $Z$, which in turn will give you the identity of the particle emitted from the nucleus of the tungsten-159 isotope.

Here $A$ represents the mass number of the particle, which as you know tells you the number of protons and electrons present in the nucleus, and $Z$ represents the atomic number, which is simply the number of protons present in the nucleus.

In any nuclear reaction, charge and mass must be conserved. This means that you can write

$159 = 155 + A \to$ conservation of mass

$\textcolor{w h i t e}{a} 74 = \textcolor{w h i t e}{a} 72 + Z \to$ conservation of charge

Solve these two equations to find

$A = 159 - 155 = 4$

$Z = 74 - 72 = 2$

You can thus say that the unknown particle will have a mass number equal to $4$ and an atomic number equal to $2$. This particle is known as an alpha particle, ""_2^4alpha, and is essentially the nucleus of a helium-4 atom.

The balanced nuclear equation will thus look like this

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{_ (color(white)(a)74)^159"W" -> ""_ (color(white)(a)72)^155"Hf" + }}_{2}^{4} \alpha} \textcolor{w h i t e}{\frac{a}{a}} |}}}$