Question

Question #187e4

Answer 1

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, `"B"`, and for the atomic number of helium, `"He"`.

Now, you know that

`""_(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 ->` conservation of mass

`color(white)(1)5 + 2 = Z + 0 ->` conservation of charge

You should end up with

`14 = A + 1 implies A = 13`

`color(white)(1)7 = Z`

The unknown element is nitrogen, `"N"`, because you have

`Z = 7 ->` the atomic number of nitrogen

The unknown nuclide is nitrogen-13 because you have

`A = 13 ->` the mass number of nitrogen-13

The complete nuclear equation will look like this

`""_ (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, `""_0^1"n"`, are produced.