Writing Alpha and Beta Decay Equations

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Here's a brief tutorial about nuclear symbols. First, an example of a nuclear symbol:

  614 C

Make sure to remember that the lower number is the atomic number and the upper number is the mass number.

atomic numbermass number element symbol

Here's another nuclear symbol:

  92238 U

That 92 is the atomic number, which is the number of protons. That 238 is the mass number, which is the sum of the protons and the neutrons.

Lastly, remember that you have to do a subtraction to get the number of neutrons:

14 − 6 = 8

238 − 92 = 146

Oh, and remember: some textbooks or teachers call it the isotopic symbol. You have to know both.

Your last, last reminder (hopefully): on Internet answer boards such as Yahoo Answers, you can't format isotopic symbols like above. Here's what I use on YA:

3-Li-7

or

Li-7 (the 3 is implied in the symbol Li)

Other people do something like this:

7/3Li
Also, really old materials (from when the ChemTeam was in school), show nuclear symbols like this:
92U235

Be prepared!


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:

  88222 Ra  --->  24 He  +    86218 Rn

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:

  84208 Po  --->    82204 Pb  +  24 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.

103256 Lr       89225 Ac       91231 Pa       87211 Fr       79185 Au

Solution:

103256 Lr  --->  101252 Md  +  24 He
 
  89225 Ac  --->    87221 Fr  +  24 He
 
  91231 Pa  --->    89227 Ac  +  24 He
 
  87211 Fr  --->    85207 At  +  24 He
 
  79185 Au  --->    77181 Ir  +  24 He

Example #4: Here are five more to try:

  92233 U       64149 Gd       90232 Th       78175 Pt       93237 Np

Solution:

  92233 U  --->  24 He  +    90229 Th
 
  64149 Gd  --->  24 He  +    62145 Sm
 
  90232 Th  --->  24 He  +    88228 Ra
 
  78175 Pt  --->  24 He  +    76171 Os
 
  93237 Np  --->  24 He  +    91233 Pa

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

  90234 Th       60144 Nd       62146 Sm       67151 Ho       78192 Pt

Solution:

  90234 Th  --->    88230 Ra
 
  60144 Nd  --->    58140 Ce
 
  62146 Sm  --->    60142 Nd
 
  67151 Ho  --->    65147 Tb
 
  78192 Pt  --->    76188 Os

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

  94239 Pu       96240 Cm       97247 Bk       99252 Es     102255 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

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:

  614 C  --->    714 N  +  −1   0 e  +  νe — 

Some points to be made about the 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:

  53131 I  --->    54131 Xe  +  −1   0 e  +  νe — 

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.

26 He     1124 Na       79201 Au     2652 Fe     1942 K

Solution:

26 He  --->  36 Li  +  −1   0 e  +  νe — 
 
1124 Na  --->  1223 Mg  +  −1   0 e  +  νe — 
 
  79201 Au  --->    80201 Hg  +  −1   0 e  +  νe — 
 
2652 Fe  --->  2752 Co  +  −1   0 e  +  νe — 
 
1942 K  --->  2042 Ca  +  −1   0 e  +  νe — 

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

3890 Sr       93239 Np       95247 Am     3582 Br       80203 Hg

Solution:

3890 Sr  --->  3990 Y  +  −1   0 e
 
  93239 Np  --->    94239 Pu  +  −1   0 e
 
  95247 Am  --->    96247 Cm  +  −1   0 e
 
3582 Br  --->  3682 Kr  +  −1   0 e
 
  80203 Hg  --->    81203 Tl  +  −1   0 e
 

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

38 Li       513 B       920 F     1532 P     4399 Tc

Solution:

38 Li  --->  48 Be
 
  513 B  --->    613 C
 
  920 F  --->  1020 Ne
 
1532 P  --->  1632 S
 
4399 Tc  --->  4499 Ru

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

  410 Be       716 N     1635 S       55137 Cs       77192 Ir

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 was 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.


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