In the last lesson, we saw that certain isotopes are more abundant than others. This occurs because some isotopes are more stable and as a result stick around longer. Since they last longer they end up comprising a larger proportion of a particular sample.
We can think of it this way, each element has a “happy place” where the number of protons and neutrons are perfectly balanced. As we add or remove neutrons an atom moves further and further away from this ideal state and becomes more and more unstable. Atoms can’t just lose protons or neutrons, with the exception of really large nuclei, because the strong force binding them together prevents them from leaving.
Instead, an atom can make one of two choices: change one of its protons into a neutron or change one of its neutrons into a proton. This is a special type of radioactive decay called β-decay.
Before delving into β-decay in more detail let’s first discuss radioactive decay as a whole. As described above certain atoms are unstable and “want” to get back to a more stable state. To do this they emit energy in the form of particles or radiation giving rise to a daughter nuclei. Daughter nuclei are simply the product of decay and result from parent nuclei giving off energy as described above.
Radioactive decay at a particle level happens randomly and due to quantum mechanics, we can’t predict what any one nuclei will do. However, when we zoom out to the sample as a whole we can accurately determine the rate of decay and represent it as a half-life. A half-life is the amount of time it takes for half of the parent nuclei in a sample to decay and be converted into daughter nuclei.
To illustrate how this works let’s look at Carbon-14, which has a half-life of 5,730 years. This means that if we have a 10-gram sample of Carbon-14 after 5,730 years only 5 grams will remain then after another 5,730 years only 2.5 grams will remain so on and so forth.
Since each successive decay halves, the amount of sample present decay ends up being exponential rather than linear. We will often be asked to calculate decay in equations so it pays to breakdown how to approach these types of questions now.
Let’s look at a step by step way of solving decay problems now by using three practice problems.
The MCAT can ask either for the amount of sample remaining or the amount of sample that has decayed. Make sure you check to see which one they want. This is important because you can walk through all the steps correctly, but end up with an answer that doesn’t actually pertain to the question.
Here 1 is asking for the starting amount, 2 asks for the amount remaining, and 3 asks for the amount decayed.
Next you will want to identify the half-life of the element in question. This information can be stated plainly in the passage itself or may be derived from a decay plot.
Here the half-life is clearly stated as 5,730 years.
After determining the half-life we will now look for the amount of time that the sample has undergone decay.
The time elapsed in all three scenarios is 26,000 years.
Once you find this time we need to convert it into a whole number of half-lives. Why a whole number? We can calculate 1.79 half-lives, but the math is challenging and since we don’t get a calculator it isn’t worth our time. The answer choices should be spaced far enough apart to estimate accurately enough.
To accomplish this we can treat the half-life as a conversion factor in the same way you would when converting from meter to kilometers.
[latexpage]
\[Carbon-14\;half-life=5,730\;years\]
\[\frac{half-life}{5730\;years}\times26,000\;years \to \frac{26,000}{6,000} \to \frac{13}{3}= 4.2ish\; half-lives\]
We will end up rounding down to 4 half-lives since that should be close enough for the MCAT
Now that we have all the information we need, we can calculate our desired value by using a couple of “equations”. We won’t be using the formal half-life equation because the math is more complicated and it becomes yet another equation we have to remember. Instead, we can use the following “equation”.
[latexpage]
\[Sample\;Remaining\;=\;\frac{1}{2^n}(Starting\;Sample)\]
Carbon-14 has a half-life of 5,730 years. A sample that is 26,000 years old gives off 3,400 millicuries of radiation, how much radiation did the sample give off initially?
[latexpage]
\[3,400\; millicuries=\;\frac{1}{2^4}x \to x = 3,400 \times 16 \to 54,400\; millicuries
\]
Carbon-14 has a half-life of 5,730 years. If an initial sample weighed 200kg how much will it weigh after 26,000 years?
[latexpage]
\[x=\;\frac{1}{2^4}\times200kg \to x = 12.5kg
\]
Carbon-14 has a half-life of 5,730 years. If an initial sample weighed 200kg how much decays after 26,000 years?
From the previous question we know that 12.5 kg remained so the rest must have decayed. Therefore…
\[200kg – 12.5kg=Decayed\;sample \to 187.5kg
\]
Half-Life | Fraction Remaining | Fraction Decayed | Percent Remaining | Percent Decayed |
1 | 1/2 | 1/2 | 50% | 50% |
2 | 1/4 | 3/4 | 25% | 75% |
3 | 1/8 | 7/8 | 12.5% | 87.5% |
n | 1/2n | 1-(1/2n) | 100/2n | 100-(100/2n) |
Alternatively you can halve the value sequentially since most questions won’t go above 5 half-lives.
Alright now that we have a big picture sense of what radioactive decay is and how to solve problems involving decay, let’s get back to β-decay.
As discussed above, atoms have two choices to become more stable. Either to change one of its neutrons into a proton or change one of its protons into a neutron. We call this specific flavor of decay β-decay because in the process of decaying the nucleus will emit a β-particle. A β-particle is just an electron (thanks physicists) and serves to balance out the charges since we can’t create or destroy charge, but must conserve it.
In β–-decay the β-particle is an electron hence the superscripted minus sign. In this type of decay, we will be converting a neutron into a proton and emitting an electron and an antineutrino (to balance the mass difference).
Since we are gaining a proton the element will change and its identity will be immediately to the right of the original element.
We can also display this change in standard nuclear notation and in order for the equation to be correct both sides must balance both charge and mass.
In β+-decay a positron is emitted instead of an electron hence the superscripted plus sign. In this type of decay, we will be converting a proton into a neutron and emitting a positron and neutrino (to balance the mass difference).
Since we are losing a proton the element will change and its identity will be immediately to the left of the original element.
We can also display this change in standard nuclear notation and in order for the equation to be correct both sides must balance both charge and mass
In turns out that nuclei have a third way of becoming more stable when shifting their protons and neutrons around called electron capture (kinda lied about this). The name pretty much sums it up here and our nuclei can gain an electron instead of losing one. Essentially, it is β–-decay in reverse. This means it will end up having the same effect as β+-decay, as seen above.
There is also another type of decay called ⍺-decay that results in the emission of an ⍺-particle, which is a helium nucleus. ⍺-decay doesn’t typically deal with isotopes because it doesn’t change the ratio of protons to neutrons instead it helps restore stability to really large nuclei.
Remember how there was a balance between the strong force and the electrostatic repulsion in an atomic nucleus and that ultimately the strong force won out. This doesn’t happen in larger nuclei because the strong force is non-existent at larger distances. So the protons on the outer surface of large nuclei go flying off.
These protons take other particles as they go because the strong force binds them to their neighbors even though it failed to bind them to the nuclei as a whole thus in ⍺-decay we will end up losing a helium nucleus or two protons and two neutrons for each step of ⍺-decay.
Since there is only one type of ⍺-decay our daughter nuclei will always be two to the left in the periodic table due to losing two protons at a time.
Again we can denote this change using standard atomic notation as follows