How do you write a nuclear equation for the alpha decay of #"_62^148Sm#?

1 Answer
Nov 13, 2016

Answer:

#""_ (color(white)(1)62)^148"Sm" -> ""_ (color(white)(1)60)^144"Nd" + ""_2^4alpha#

Explanation:

The thing to remember about alpha decay is that it occurs when the nucleus of a radioactive nuclide emits an alpha particle, #alpha#, which is essentially the nucleus of a helium-4 atom.

https://www.mirion.com/introduction-to-radiation-safety/types-of-ionizing-radiation/

Simply put, an alpha particle contains #2# protons and #2# neutrons, which implies that it has a mass number equal to #4#.

Therefore, you can use isotopic notation to write the alpha particle using its atomic number of #2# and mass number of #4#

#""_2^4alpha#

You can now set up the nuclear equation that describes the alpha decay of samarium-148

#""_ (color(white)(1)color(blue)(62))^color(orange)(148)"Sm" -> ""_color(blue)(Z)^color(orange)(A)"X" + ""_color(blue)(2)^color(orange)(4)alpha#

In order to find the identity of the daughter nuclide, use the fact that mass and charge are conserved in a nuclear equation

#color(orange)(148 = A + 4)" " -># conservation of mass

#color(white)(1)color(blue)(62 = Z + 2)" " -># conservation of charge

Solve to find the values of #A# and #Z#

#148 = A + 4 implies A = 144#

#color(white)(1)62 = Z + 2 implies Z = 60#

Grab a periodic table and look for the element which has the atomic number equal to #60#. This element is neodymium, #"Nd"#. The daughter nuclide is neodybium-144.

The balanced nuclear equation that describes the alpha decay of samarium-148 will thus be

#color(darkgreen)(ul(color(black)(""_ (color(white)(1)62)^148"Sm" -> ""_ (color(white)(1)60)^144"Nd" + ""_2^4alpha)))#