How do you write a balanced nuclear equation for alpha decay of Po-218?

Jul 11, 2017

Here's how you can do that.

Explanation:

When a radioactive nuclide undergoes alpha decay, it emits an alpha particle, ""_2^4alpha, which is essentially the nucleus of a helium-4 atom.

This means that after the alpha particle is emitted

• the mass number of the nuclide will decrease by $4$ $\to$ this happens because the alpha particle contains $2$ protons and $2$ neutrons
• the atomic number of the nuclide will decrease by $2$ $\to$ this happens because the alpha particle contains $2$ protons

You can thus say that you have

${\text{_ (color(white)(.)84)^218"Po" -> ""_ (color(white)(.)(84-2))^((218 - 4))"X" + }}_{2}^{4} \alpha$

A quick look in the Periodic Table of Elements will show that the element that has the atomic number equal to

$84 - 2 = 82 \to$ conservation of charge

is lead, $\text{Pb}$. This means that the resulting nuclide will be lead-214 since its mass number is equal to

$218 - 4 = 214 \to$ conservation of mass

The balanced nuclear equation that describes the alpha decay of polonium-218 will look like this

${\text{_ (color(white)(.)84)^218"Po" -> ""_ (color(white)(.)82)^214"X" + }}_{2}^{4} \alpha$

As