# Question #187e4

Jul 29, 2017

The unknown nuclide is nitrogen-13.

#### Explanation:

The thing to keep in mind about nuclear equations is that mass and charge must be conserved.

Before doing anything else, grab a Periodic Table and look for the atomic number of boron, $\text{B}$, and for the atomic number of helium, $\text{He}$.

Now, you know that

$\text{_(color(white)(1)5)^10"B" + ""_2^4"He" -> ""_Z^A? + ""_0^1"n}$

In order to find the identity of the unknown nuclide, you must take into account that

$10 + 4 = A + 1 \to$ conservation of mass

$\textcolor{w h i t e}{1} 5 + 2 = Z + 0 \to$ conservation of charge

You should end up with

$14 = A + 1 \implies A = 13$

$\textcolor{w h i t e}{1} 7 = Z$

The unknown element is nitrogen, $\text{N}$, because you have

$Z = 7 \to$ the atomic number of nitrogen

The unknown nuclide is nitrogen-13 because you have

$A = 13 \to$ the mass number of nitrogen-13

The complete nuclear equation will look like this

$\text{_ (color(white)(1)5)^10"B" + ""_ 2^4"He" -> ""_ (color(white)(1)7)^13"N" + ""_0^1"n}$

According to this nuclear equation, when a boron-10 nucleus is bombarded with a helium-4 nucleus, also known as an alpha particle, $\alpha$, a nitrogen-13 nucleus and a neutron, $\text{_0^1"n}$, are produced.