Question #187e4

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.