Question

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?

Answer 1

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.

![http://www.ck12.org/book/CK-12-Physical-Science-Concepts-For-Middle-School/section/3.60/]()

Pull up a periodic table and look for the atomic number of samarium, `"Sm"`. You'll find that it's equal to `62`. This means that you can start by writing

`""_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.

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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 ->` conservation of mass number**

`62 = x + 2 ->` conservation of charge**

You will thus get

`{(y = 149 - 4 = 145), (x = 62 -2 = 60) :}`

The element that has an atomic number equal to `60` is neodymium, `"Nd"`.

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

`""_62^149"Sm" -> ""_60^145"Nd" + ""_2^4"He"`