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

##### 1 Answer

#### Explanation:

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

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

Therefore, you can use isotopic notation to write the alpha particle using its atomic number of

#""_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

#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 *neodymium*,

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)))#