# What are the products for the alpha decay of ""_64^150Gd?

Jan 13, 2016

Here's what I got.

#### Explanation:

Alpha decay is said to take place when a radioactive nuclide decays by emission of an $\alpha$ particle, which is an alternative term used to denote the nucleus of a helium-4 atom, $\text{_2^4"He}$. So right from the start, you know that one of the products will be an $\alpha$ particle.

$\text{_(color(white)(x)64)^150"Gd" -> color(blue)(""_x^y?) + ""_2^4"He}$

In order to determine the identity of that color(blue)(""_x^y?) isotope, you need to take into account the fact that nuclear equations must conserve, among other things, charge and mass number.

Mass number, which represents the number of protons and neutrons located in the nucleus, is written in the upper-left of the chemical symbol.

The atomic number, which tells you the number of protons located in the nucleus, is written in the lower-left of the chemical element. So, looking at your nuclear equation, you can say that

$150 = y + 4 \to$ conservation of the mass number

and

$64 = x + 2 \to$ conservation of charge

This will give you

$\left\{\begin{matrix}y = 150 - 4 = 146 \\ x = 64 - 2 = 62\end{matrix}\right.$

A quick look in the periodic table will reveal that samarium, $\text{Sm}$, has an atomic number equal to $62$. This means that the second product of the decay will be the samarium-156 isotope

$\text{_(color(white)(a)64)^150"Gd" -> ""_62^146"Sm" + ""_2^4"He}$