Question #187e4
1 Answer
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,
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,
#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,