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

Nov 13, 2016

${\text{_ (color(white)(1)62)^148"Sm" -> ""_ (color(white)(1)60)^144"Nd" + }}_{2}^{4} \alpha$

#### 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.

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

${\text{_ (color(white)(1)color(blue)(62))^color(orange)(148)"Sm" -> ""_color(blue)(Z)^color(orange)(A)"X" + }}_{\textcolor{b l u e}{2}}^{\textcolor{\mathmr{and} a n \ge}{4}} \alpha$

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

$\textcolor{\mathmr{and} a n \ge}{148 = A + 4} \text{ } \to$ conservation of mass

$\textcolor{w h i t e}{1} \textcolor{b l u e}{62 = Z + 2} \text{ } \to$ conservation of charge

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

$148 = A + 4 \implies A = 144$

$\textcolor{w h i t e}{1} 62 = Z + 2 \implies Z = 60$

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

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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{_ (color(white)(1)62)^148"Sm" -> ""_ (color(white)(1)60)^144"Nd" + }}_{2}^{4} \alpha}}}$