### Writing Alpha and Beta Decay Equations

Alpha Decay In 1899, Ernest Rutherford wrote the following words:

"These experiments show that the uranium radiation is complex and that there are present at least two distinct types of radiation - one that is very readily absorbed, which will be termed for convenience the alpha-radiation, and the other of more penetrative character which will be termed the beta-radiation."

The image to the right is of a twenty-eight year old Ernest Rutherford while at McGill University in 1899.

Alpha decay can most simply be described like this:

1) The nucleus of an atom splits into two parts.
2) One of these parts (the alpha particle) goes zooming off into space.
3) The nucleus left behind has its atomic number reduced by 2 and its mass number reduced by 4 (that is, by 2 protons and 2 neutrons).

There are other points, but the three above are enough for a start.

Example #1: A typical alpha decay equation:

 ---> ${\text{}}_{2}^{4}\text{He}$ +

Notice several things:

1) The atom on the left side will be the one that splits into two pieces.
2) One of the two atoms on the right will ALWAYS an alpha particle.
3) The other atom on the right ALWAYS will go down by two in the atomic number and four in the mass number.

And, notice one more thing. Never put the alpha particle on the left-hand side. It ALWAYS goes on the right-hand side.

Example #2: Another alpha decay:

 ---> + ${\text{}}_{2}^{4}\text{He}$

Check it and compare the three points to the example. Keep in mind that this equation shows the left-hand side splitting into the two pieces shown on the right-hand side.

Notice how I wrote the alpha particle as the first product in Example #1, but as the second product in Example #2. Some teachers insist on it going second. It's a stylistic thing, with no widely-accepted standard as to where the alpha particle goes.

Example #3: Write the alpha decay equations for these five nuclides.

 ${\text{}}_{103}^{256}\text{Lr}$

Solution:

 ${\text{}}_{103}^{256}\text{Lr}$ ---> ${\text{}}_{101}^{252}\text{Md}$ + ${\text{}}_{2}^{4}\text{He}$ ---> + ${\text{}}_{2}^{4}\text{He}$ ---> + ${\text{}}_{2}^{4}\text{He}$ ---> + ${\text{}}_{2}^{4}\text{He}$ ---> + ${\text{}}_{2}^{4}\text{He}$

Example #4: Here are five more to try:

Solution:

 ---> ${\text{}}_{2}^{4}\text{He}$ + ---> ${\text{}}_{2}^{4}\text{He}$ + ---> ${\text{}}_{2}^{4}\text{He}$ + ---> ${\text{}}_{2}^{4}\text{He}$ + ---> ${\text{}}_{2}^{4}\text{He}$ +

Example #5: Show only the daughter nuclide on these last five:

Solution:

 ---> ---> ---> ---> --->

Example #6: Here are five more, but no answers.

 ${\text{}}_{102}^{255}\text{No}$

Example #7: The element 85-At-213 decays to 83-Bi-209. (a) Why type of decay is this? (b) Write the nuclear equation for the decay.

Solution for (a):

Notice how the atomic number went down by 2 and the mass number went down by 4.

What nuclide has an atomic number of 2 and a mass number of 4?

This: 2-He-4.

That's a helium nucleus (also called an alpha particle) and it is associated with alpha decay.

Solution for (b):

85-At-213 ---> 83-Bi-209 + 2-He-4

Example #8: Write the equation for the alpha decay of Po-218 two different ways.

Solution:

1) Here is the official way, using superscripts and subscripts:

 ---> + ${\text{}}_{2}^{4}\text{He}$

2) Here it is, writing everything on the same line:

84-Po-218 ---> 82-Pb-214 + 2-He-4

3) Here's a third way, also with everything on one line:

218/84Po ---> 214/82Pb + 4/2He

Personally, I don't like the third way, but it does show up often out there on ye olde Internet.

Beta Decay

Beta decay is somewhat more complex than alpha decay is. These points present a simplified view of what beta decay actually is:

1) A neutron inside the nucleus of an atom breaks down, changing into a proton.
2) It emits an electron and an antineutrino (more on this later), both of which go zooming off into space.
3) The atomic number goes UP by one and mass number remains unchanged.

Example #1: A beta decay equation:

 ---> + +

1) The nuclide that decays is the one on the left-hand side of the equation.
2) The order of the nuclides on the right-hand side can be in any order.
3) The way it is written above is the usual way.
4) The mass number and atomic number of the antineutrino are zero and are, in modern usage, not written.
5) The bar above the symbol indicates it is an anti-particle.
5) The neutrino symbol is the Greek letter "nu."

Example #2: Another example:

 ---> + +

Notice that all the atomic numbers on both sides ADD UP TO THE SAME VALUE and the same for the mass numbers.

By the way, an older style for the antineutrino symbol adds on two zeros where the atomic number and the mass number are placed, as well as dropping the subscripted e. I couldn't make the formatting work, so I have to describe it in words. You might wind up with an older teacher who insists on the older style of writing the antineutrino. Or, you might be using an older set of materials.

Example #3: Write out the full beta decay equation for each of the five.

 ${\text{}}_{2}^{6}\text{He}$ ${\text{}}_{11}^{24}\text{Na}$ ${\text{}}_{26}^{52}\text{Fe}$ ${\text{}}_{19}^{42}\text{K}$

Solution:

 ${\text{}}_{2}^{6}\text{He}$ ---> ${\text{}}_{3}^{6}\text{Li}$ + + ${\text{}}_{11}^{24}\text{Na}$ ---> ${\text{}}_{12}^{23}\text{Mg}$ + + ---> + + ${\text{}}_{26}^{52}\text{Fe}$ ---> ${\text{}}_{27}^{52}\text{Co}$ + + ${\text{}}_{19}^{42}\text{K}$ ---> ${\text{}}_{20}^{42}\text{Ca}$ + +

Example #4: Some teachers do not want the antineutrino included. Be sure to do what your teacher wants.

 ${\text{}}_{38}^{90}\text{Sr}$ ${\text{}}_{35}^{82}\text{Br}$

Solution:

 ${\text{}}_{38}^{90}\text{Sr}$ ---> ${\text{}}_{39}^{90}\text{Y}$ + ---> + ---> + ${\text{}}_{35}^{82}\text{Br}$ ---> ${\text{}}_{36}^{82}\text{Kr}$ + ---> +

Example #5: Five more with just the daughter nuclide:

 ${\text{}}_{3}^{8}\text{Li}$ ${\text{}}_{15}^{32}\text{P}$ ${\text{}}_{43}^{99}\text{Tc}$

Solution:

 ${\text{}}_{3}^{8}\text{Li}$ ---> ${\text{}}_{4}^{8}\text{Be}$ ---> ---> ${\text{}}_{10}^{20}\text{Ne}$ ${\text{}}_{15}^{32}\text{P}$ ---> ${\text{}}_{16}^{32}\text{S}$ ${\text{}}_{43}^{99}\text{Tc}$ ---> ${\text{}}_{44}^{99}\text{Ru}$

Example #6: Here are five more, but no answers.

 ${\text{}}_{16}^{35}\text{S}$

Example #7: Cm-247 decays into a more stable isotopes by emitting 3 alpha particles followed by 2 beta particles. What nuclide results from these decays?

Solution:

1) Step-by-step, describe three alpha decays:

96-Cm-247 loses an alpha (which is 2-He-4) and becomes 94-Pu-243 (see how it went down by 2 in the atomic number and down by 4 in the mass number?)

94-Pu-243 loses 2-He-4 and becomes 92-U-239

92-U-239 loses 2-He-4 and becomes 90-Th-235

2) When you lose a beta (remember, this is an electron), the atomic number goes up by 1 and the mass number does not change.

90-Th-235 ---> 91-Pa-235

91-Pa-235 ---> 92-U-235

3) Let's write it as one complete equation:

--->   ${\text{}}_{Z}^{A}\text{X}$   +   3 ${\text{}}_{2}^{4}\text{He}$   +   2

atomic number ---> 96 − (6 + (−2)) = 92
mass number ---> 247 − (12 + 0) = 235

${\text{}}_{Z}^{A}\text{X}$   is

A Brief Note on the Antineutrino

As beta decay was studied over the years following 1899, it was found that the same exact beta decay produced an electron with variable energies.

For example, let us study Li-8 becoming Be-8. Each atom of Li-8 produces an electron and the theory says all the electrons should have the same energy.

This was not the case.

The electrons were coming out with any old value they pleased up to a maximun value, characteristic of each specific decay.

To make a long story short, Wolfgang Pauli (in about 1930 or so) suggested the energy was being split randomly between two particles - the electron and an unknown light particle that was escaping detection. Enrico Fermi suggested the name "neutrino," which is Italian for "little neutral one."

The neutrino itself was not detected until 1956 and the discoverers informed Pauli just a few months before his death due to cancer. Later on, it was discovered that it was an antineutrino that was produced in beta decay.