1 RANDOMNESS AND RADIOACTIVE DECAYS

Radioactivity and Half Lives

1 Randomness and radioactive decays

Quantum mechanics does not typically predict definite outcomes: it gives usa range of possibilities, and lets us calculate the probability of each possibility.Individual events, then, are not predictable beyond these probabilities. Butif we have lots of events of the same type, we can be confident about ourpredictions about statistical quantities such as averages.

A classic example is the radioactive decay of large and unstable atomic nu-clei. Small atomic nuclei—those with just a handful of protons and neutrons—are strongly bound and stable. Indeed, there is an attraction between protonsand neutrons that is known as the strong nuclear force or the strong forcefor short. Physicists speak of the four fundamental forces in nature: grav-ity, electromagnetism, and the strong and weak nuclear forces. The strongforce is not an inverse-square force; it falls off much more rapidly, and is notnoticeable beyond nuclear distances ∼ 10−15 m. But closer than that, it isextremely strong—strong enough to totally overwhelm the otherwise quitestrong electrical repulsion between the protons packed closely together inatomic nuclei.

Small nuclei are stable, and as they get larger, the protons and neutronsbecome ever more tightly bound. So if you fuse very small nuclei to produceslightly larger nuclei, you release energy. This process of fusion is the basis forthe energy output of stars such as our Sun. In the incredibly hot temperaturesat the cores of stars, Hydrogen atoms come together to form Helium, releasingenergy. Every chemical element in the periodic table up to iron (Fe) is formedby fusion in stars this way.

Beyond iron, however, it costs energy to make nuclei larger, and far downin the periodic table, elements become increasingly unstable. Then you canrelease energy by fission, when you break up a large nucleus into smallerpieces. Elements heavier than iron don’t form in stars but in the violent

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2 AN EXPERIMENT

deaths of stars. When a star runs out of fuel—elements lighter than iron tofuse—it can explode as a supernova, and the shock wave that results is soviolent that it provides the extra energy needed to force nuclei to fuse intoelements heavier than iron.

The most common form of breakup of large nuclei you will encounter isradioactive decay, where a large unstable nucleus becomes smaller by shoot-ing out highly energetic particles. These are usually called alpha (α; Henuclei, two protons and two neutrons bound together), beta (β; electrons),and gamma (γ; very high frequency photons).

Radioactive decays are quantum events. Therefore when each individualnucleus will decay is not exactly predictable, we can only figure out a prob-ability. The customary way of expressing this is to give the time in whichany individual radioactive nucleus has a 1

2chance of decaying: this is the

half-life, t1/2.

2 An experiment

Let’s do an experiment that simulates radioactive decays. All you will needis a coin you can flip.

You will have a starting population of 18 identical radioactive nuclei. Flipa coin for each one. If you get heads, record a checkmark in the first line ofthe table: it has survived the first t1/2. If you get tails, don’t put anythingdown: that nucleus has decayed and become something else, it’s no longerpart of your sample and you will no longer be flipping a coin for it. At theend of the row, record the total number of checkmarks; that is your survivingpopulation number. Then flip the coin once again for each surviving nucleus,and record these results in the second row. Keep going until you have filledthe rows for five half-lives.

t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Total

t1/22t1/23t1/24t1/25t1/2

You will notice that after each half-life, it is not exactly half the pop-

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2 AN EXPERIMENT

ulation that will survive. There are statistical fluctuations, usually aboutthe size of one standard deviation or less, above and below the populationhalfway mark. Here’s a useful rule of thumb: the size of the average pop-ulation is ∝ N , where N is your starting population size. In our case, onaverage, N/2 will survive the passing of a a half-life. Since you start with18, you should end up around 9, then around 4.5, and so on. The size of thefluctuations, on the other hand, should be ∝

√N . 18 will not go down to 9

exactly, but most of the time, down to within 9±√

9 = 9± 3.Now repeat the experiment two more times:

t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Total

t1/22t1/23t1/24t1/25t1/2

t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Total

t1/22t1/23t1/24t1/25t1/2

Notice that the outcomes are similar but not the same, and that thehistory of each individual nucleus in each experiment can be very different.

As the next step, I want you to sketch a graph of how the total populationchanges in all three experiments, on the same graph. Then also draw in athick line representing the expected (average) total population values, goingthrough 18, 9, 4.5, and so on. Finally, draw in two dotted lines aboe andbelow the average line that indicate the range most fluctuations will fall in,going through 18±

√18 and 9±

√9 and so on.

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2 AN EXPERIMENT

This would be what your graph looked like. And if you were rerun theexperiment lots of times, not just three, your graph would a mess where thedarkest part coincided with the average line, and got less intense in color asyou moved away from average. I’ve tried to indicate this by the degree of redshading—the average values are the most probable, but you will typically bea little off, and there will be rare fluctuations that will take the experimentfar off the most probable line.

Now, we won’t do it, but if your starting population was not 18 but 1800,your graphs, when scaled to the same size on paper, would end up quitesimilar. But since the average goes like N and the standard deviation like√N , the dark part of your graph would be less thick: the relative size of the

statistical fluctuation would go like√N/N = 1/

√N , which gets smaller as

N increases. This is important: the fluctuations still grow in size, but theygrow much slower than the expected average values. The relative size goingdown tells you that the expected, average result becomes a better and betterbet as the population increases.

In your first graph, at the first half-life, the fluctuations were mostlywithin 9±

√9 = 9±3. But now, after the first half-life, the average population

will be 900, and the fluctuations will fall in 900±√

900 = 900± 30. 30 is alot smaller compared to 900, not like 3 compared to 9.

Say the thickness of your line (the dark area) on paper in your N = 18graph started out around 1 cm. With N = 1800 = 18×102, the thickness willbe reduced to 1/

√102 = 0.1 cm, still thickish. If you then did N = 18× 106,

just 18 million, the thickness would be 1/√

106 = 0.001 cm, very thin. Youcan see this in the following graphs. With the population in the millions,

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2 AN EXPERIMENT

the graph on the right just has a thin line, and the average behavior is justabout guaranteed.

The number of particles in everyday objects is typically around 1024. Sayyou started with N = 18 × 1024. What would the thickness of the line ofyour graph now be? Compare it to the size of an atom, around 10−10 m.

When you find this result, also ask yourself: quantum mechanics is ourcurrent most fundamental theory of physics. Individual particles, such asradioactive nuclei, behave randomly. Everything we see around us is madeout of such particles, and all physical processes involve interactions of suchparticles. And yet, everyday life is not full of events that happen randomly,without cause or pattern. You’ve had a year of physics, and you have seenboatloads of equations that tell you that if you measure some variables pre-cisely, you can determine the exact values of something else. There hasn’tbeen much randomness in physics until now. So how do we get a nonrandommacroscopic world out of a microscopic world where the physics is random?All this messing around with graphs is supposed to help you understand how.

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3 HALF-LIVES AND EXPONENTIALS

The thickness of your line would be 1/√

1024 = 10−12 cm, which is muchsmaller than 10−8 cm for an atom. The expected (average) behavior will hap-pen with near-certainty. You have no hope of seeing statistical fluctuations.That is, by the way, why the macroscopic world looks so non-random. Withenormous populations of particles, fluctuations become negligible and we canstake our lives (and we do) on equations that describe the average behavior.

3 Half-lives and exponentials

The shape of the expected population curves is a typical exponential de-cay, exactly like that for a discharging capacitor. The equation for how thepopulation of radioactive nuclei in a sample changes with time is

N(t) = N(0)(

1

2

)t/t1/2

Quantities that are proportional to N behave the same way. The mass of theradioactive part of the sample, m, for example; or the activity of the sample(how strongly radioactive it is) which you detect with a Geiger counter:

m(t) = m(0)(

1

2

)t/t1/2

, a(t) = a(0)(

1

2

)t/t1/2

These are exponentials. After all,(1

2

)t/t1/2

= 2−t/t1/2 =(eln 2

)−t/t1/2= e−(ln 2) t/t1/2

Radioactivity exhibits exponential behavior like a lot of other things inphysics. The reason is simple. Whenever the rate of change of a quantityis proportional to the quantity itself, exponential growth or decay happens.For example, in the early stages of a pandemic, when the infected populationis small and therefore no resource limitations such as a finite host populationare yet in sight, the infection rate is proportional to the number of peoplealready infected. Hence you get exponential growth.

Let’s finish with a simple calculation. You have a radioactive sample thatgives you an activity of 3.16 counts/second on your Geiger counter. Exactly12 days later, you measure an activity of 2.87 counts/second. Assuming nocomplications such as multiple types of decay going on simultaneously, whatis the half-life of your sample?

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3 HALF-LIVES AND EXPONENTIALS

Just put the numbers into the exponential decay expression.

a(t) = a(0)(

1

2

)t/t1/2

⇒ ln

[a(t)

a(0)

]= −(ln 2)

t

t1/2

⇒ t1/2 = − ln 2

ln [a(t)/a(0)]t

In our case,

t1/2 = − ln 2

ln (2.87/3.16)12 = 86.4 days

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