# It takes about 10^16 years for just half the samarium-149 in nature to decay by alpha-particle emission. What is the decay equation, and what is the isotope that is produced by the reaction?

Jan 15, 2016

The alpha decay of samarium-149 produces neodymium-145.

#### Explanation:

When a radioactive isotope undergoes alpha decay, its nucleus emits an $\alpha$ particle, which is a term used to denote the nucleus of a helium-4 atom. Pull up a periodic table and look for the atomic number of samarium, $\text{Sm}$. You'll find that it's equal to $62$. This means that you can start by writing

$\text{_62^149"Sm" -> ""_x^ycolor(blue)(?) + ""_2^4"He}$

Your goal now is to find the values of $x$ and $y$ by using the fact that nuclear equations must conserve, among other things, charge and mass number.

At this point, it's important that you're familiar with isotope notation. For samarium-149, you have an atomic number equal to $62$ and a mass number equal to $149$. This means that you can write

$149 = y + 4 \to$ conservation of mass number

$62 = x + 2 \to$ conservation of charge

You will thus get

$\left\{\begin{matrix}y = 149 - 4 = 145 \\ x = 62 - 2 = 60\end{matrix}\right.$

The element that has an atomic number equal to $60$ is neodymium, $\text{Nd}$.

The alpha decay of samarium-149 will thus produce neodymium-145 and an alpha particle.

$\text{_62^149"Sm" -> ""_60^145"Nd" + ""_2^4"He}$