How can I solve nuclear equations?

1 Answer
May 30, 2014

Nuclear equations can be solved quite simply: let's do one example of alpha decay and one of beta decay.


First, a quick revision of radioactive decay:

During alpha decay, an alpha particle is emitted from the nucleus —- it is the equivalent of a helium atom (i.e. it has a mass of 4 and an atomic number of 2). So, let's take the following question:

Polonium-210 is a radioisotope that decays by alpha-emission. Write a balanced nuclear equation for the alpha decay of polonium-210.

In symbols, the equation becomes

#""_84^210"Po" → ? +color(white)(l) _2^4"He"#

The sums of the superscripts and of the subscripts must be the same on each side of the equation.

Take 4 away from the mass number (210-4 = 206)

Take 2 away from the atomic number (84-2 = 82). Lead is element number 82.

So, the equation is

#""_84^210"Po" → _82^206"Pb" +color(white)(l) _2^4"He"#

Now let's try one for beta decay — remember that, in beta decay, a neutron turns into a proton and emits an electron from the nucleus (we call this a beta particle)

Write a balanced nuclear equation for the beta decay of cerium-144)

In nuclear equations, we write an electron as #color(white)(l)_text(-1)^0"e"#. The equation becomes

#""_58^144"Ce" " → "?"" + color(white)(l) _text(-1)^0"e"#

Add one to the atomic number (58+1 = 59).
Don't change the mass number
Praseodymium is element 59
The answer is Pr-144.

#""_58^144"Ce" → _59^144"Pr" + _text(-1)^0"e"#

Here's a fission reaction.

A nucleus of uranium-235 absorbs a neutron and splits in a chain reaction to form lanthanum-145, another product, and three neutrons. What is the other product?

We write a neutron as #""_0^1"n"#, so the equation is

#""_92^235"U" + _0^1"n" → _57^145"La" + "X" + 3 _0^1"n"#

Sum of superscripts on left = 236. Sum of superscripts on right = 148. So #"X"# must have mass number = 236 – 148 = 88.

Sum of subscripts on left = 92. Sum of subscripts on right = 57. So #"X"# must have atomic number = 92 – 57 = 35. Element 35 is bromine.

The nuclear equation is

#""_92^235"U" + _0^1"n" → _57^145"La" + _35^88"Br" + 3 _0^1"n"#