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?

1 Answer
Jan 15, 2016

Answer:

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.

http://chemkjj.blogspot.ro/2015/06/c1-12-proton-number-mass-number-ions.html

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"#