14.5 Draft: Radioactivity

Nuclear decay is governed by chance. It is impossible to tell exactly when any specific nucleus will decay. Therefore, the decay is phrased in terms of statistical quantities like specific decay rate, lifetime and half-life. This section explains how these are defined.

14.5.1 Draft: Half-life and decay rate

As a generic example of an unstable nucleus, consider tritium, an isotope of hydrogen. The ${}\fourIdx{3}{1}{}{}{\rm {H}}$ tritium nucleus, the triton, consists of one proton and two neutrons. The triton suffers beta decay. Eventually it will eject an electron and an antineutrino. This turns one of the two neutrons into a proton. So the decay turns the triton into the ${}\fourIdx{3}{2}{}{}{\rm {He}}$ helium nucleus isotope called the helion. The original triton is lost.

Unlike the normal ${}\fourIdx{4}{2}{}{}{\rm {He}}$ helium nucleus, the helion contains only one neutron. However, it is stable; there will not be any further spontaneous decays.

(Note that the triton decays even though it has two neutrons, a magic number. But the helion it decays to has two protons, also magic. And just like a lone neutron decays into a less heavy proton, two neutrons in the tiny triton is just too much of a neutron excess compared to the negligible additional Coulomb repulsion in the helion.)

The big question in this section is, when will an unstable nucleus like the triton decay? Unfortunately, there is no complete anwer to that question. A given triton might last for 10 years, or it might last for 20 years or whatever. It could last less than a year, though that is not very likely. It could last for 100 years, even though that is much less likely still. But there is no way to tell for sure.

However, suppose you take a very large number of tritons and record for each how long the triton lives. Then you can average all these times and you get a number that is called the “lifetime” $\tau$ of the triton. The correct term would be expected or average lifetime, but this is physics, remember. Correct terms are not allowed. (To be fair, some physicists do use the proper term “mean lifetime” instead of just lifetime.)

If you average over enough tritons, you will find this mean lifetime of tritons to be almost 18 years. But not a single triton will decay after exactly 18 years. It is much like the expected lifetime of a newborn baby in the USA is, say, 80 years. Despite that, almost no one dies on their 80th birthday. Some die at birth or as kids.

Still, there is one big difference between people and nuclei. If you have a person who is 80 years old, surely you do not expect them to live until they become 160 years old. But if you have a bunch of tritons, on average this bunch of tritons will last for another 18 years. That is regardless of how long these tritons have survived already when you start observing them. Nuclear decay is a completely random process that occurs out of the blue; it does not depend on any previous history of the nucleus.

There is another issue. Unless you are an accountant by calling, why would you want to sit down, measure lifetimes of nuclei, and average them? What is the use?

A physically much more relevant scenario is that you have managed to create a large number of tritons, and you would like to know how long they will last for doing experiments. In particular, you might want to know how long it takes before half of the tritons you created with blood, tears, and tax-payer money, are gone. This physically more meaningful time period is called the half-life $\tau_{1/2}$ . It is related to the mean lifetime by a simple factor $\ln2$ :

$\begin{displaymath} \fbox{$\displaystyle \tau_{1/2} = \tau \ln 2 $} \end{displaymath}$

(14.3)

The half-life of the triton is 12.32 years. So if you initially have a large collection of tritium nuclei, after 12.32 years only half will be left. After 24.64 years, only a quarter will remain, and after a century only 0.4%. After a millennium only 4 10 $\POW9,{-23}$ % will remain. (Since a gram of tritium contains about 2 10 $\POW9,{23}$ atoms, after a millenium, not a single atom would be left of a gram of tritium, if you managed to create that many of them!)

Now the earth is over 4 million millenia old. So you will appreciate that almost none of the tritium ever present on earth still exists. Some new tritium is continuously being created in the atmosphere by high-energy cosmic rays, but because of the geologically short half-life, there is no measurable accumulation. The total amount of tritium present on earth is virtually zero.

**Figure 14.3:** Nuclear half-lifes. [pdf][con]
$\begin{figure}\centering \setlength{\unitlength}{1pt} \begin{picture}(405,56... ...ebox(0,0)[l]{$\fourIdx{209}{83}{}{}{\rm {Bi}}$}} } \end{picture} \end{figure}$

As in an earlier section, figure 14.3 shows again the decay processes of the nuclei. But unlike in the earlier figure 14.2, this time the square size of each nucleus has been adjusted to illustrate its half-life.

For the full size squares in the figure, the half-life is 10 $\POW9,{18}$ seconds or longer. Now 10 $\POW9,{18}$ seconds is about twice the estimated age of the universe since the Big Bang. So for nuclei that really have full size squares in figure 14.3, most of these nuclei that the universe ever created is likely to be still around.

On the other hand, if the square size of a nucleus is even slightly smaller than full size, then most of these nuclei that the universe ever created will have decayed. You may note that the square size of the ${}\fourIdx{3}{1}{}{}{\rm {H}}$ triton is noticably smaller than full size.

On the other hand, all stable green nuclei have full size squares. You might also note bismuth-209, ${}\fourIdx{209}{83}{}{}{\rm {Bi}}$ . While not actually stable, for all practical puposes it is. Its half-life of 6 10 $\POW9,{26}$ seconds exceeds the age of the unverse by a factor of over a billion. In fact, it took physicists until 2003 to observe that bismuth-209 did actually suffer alpha decay; until then it was believed stable. (Note that bismuth 209 has 126 neutrons, a magic number.) The various double-beta-decay light green nuclei live even longer, on the order of 10 $\POW9,{30}$ seconds.

Based on various arguments, it was decided to take the minimum half-life shown in figure 14.3 to be one nanosecond. Clearly the figure needs some lower limit. And a nanosecond is really fast for alpha decay, and much faster than any beta decay. (Note that you do not see any really small red or blue squares in figure 14.3, and only a few yellow ones.) And for gamma decay, a nanosecond is often used as a cut-off between prompt” and “isomeric decay.

Still, some nuclear decay processes are much quicker than a nanosecond. For example, you might note that the light nuclei that decay though proton or neutron emission in figure 14.2 have disappeared in figure 14.3. Such a decay process may have a half life on the order of 100 ys, (i.e. 100 10 $\POW9,{-24}$ seconds). That is much faster than the normal decays. In fact, in terms of quasi-classical physics, these decay times are comparable to the time for a nucleon with a speed of say a tenth of that of light to move just once through a nucleus with a size of the order of femtometers.

Another notable nucleus that has disappeared is ${}\fourIdx{8}{4}{}{}{\rm {Be}}$ , beryllium-8. Berylium-8 falls apart in two ${}\fourIdx{4}{2}{}{}{\rm {He}}$ , helium-4, alpha particles. The half-life of that process is just 67 as (i.e. 67 10 $\POW9,{-18}$ seconds). While it is technically called alpha-decay because an alpha particle is emitted, it is physically very different from the normal alpha decay of heavy nuclei. In the alpha decay of heavy nuclei, a heavy nucleus is left, not a second alpha particle. In particular, the berylium-8 decay is many orders of magnitude faster than the normal alpha decays discussed in section 14.11.

Besides (mean) lifetime and half-life, there is one more related term that is commonly used in describing nuclear decays. It is called the “specific decay rate” $\lambda$ . The specific decay rate is the relative amount of nuclei in a large sample that is lost per unit time. Mathematically, the specific decay rate is just the inverse of the mean lifetime $\tau$ :

$\begin{displaymath} \fbox{$\displaystyle \lambda \equiv \frac{1}{\tau} $} \end{displaymath}$

(14.4)

To better understand the various variables mathematically, it may be worthwhile to see how the mentioned relationships between them arise. First, according to the very definition of the decay rate $\lambda$ above, if the current number of nuclei is

, then the number of nuclei that are lost, call it $-{\rm d}I$ , in an infinitesimally small time interval ${\rm d}{t}$ is given by

$\begin{displaymath} - {\rm d}I = \lambda I { \rm d}t \end{displaymath}$

(14.5)

$\begin{displaymath} I = I_0 e^{-\lambda t} \end{displaymath}$

(14.6)

$\begin{displaymath} \tau=\left.\int_0^{\infty}t\frac{-{\rm d}I}{{\rm d}t}{ \rm d}t \right/ \int_0^{\infty}\frac{-{\rm d}I}{{\rm d}t}{ \rm d}t \end{displaymath}$

14.5.2 Draft: More than one decay process

One very important point must be emphasized. Many nuclei undergo more than one decay process. In that case, each decay process has its own decay rate, independent of the other decay processes. In such cases,

14.5.3 Draft: Other definitions

You probably think that having three different names, the specific decay rate $\lambda$ , the lifetime $\tau$ , and the half-life $\tau_{1/2}$ , for essentially the same physical quantity, is no good. You want more! Physicists are only too happy to oblige. How about using the term “decay constant” instead of specific decay rate? Its redeeming feature is that constant is a much more vague term, maximizing confusion. Even better, how does “disintegration constant” sound? Especially since the nucleus clearly does not disintegrate in decays other than spontaneous fission? Why not call it “specific activity,”come to think of it? Activity is another of these vague terms that the hated nonspecialists cannot make heads or tails of.

How about calling the product $\lambda{I}$ the “decay rate” or“disintegration rate” or simply the “activity?”

You probably want some units to go with that! What is more logical than to take the decay rate or activity to be in units of “curie,” with symbol Ci and equal 3.7 10 $\POW9,{10}$ decays per second. (Of course you guessed that straight away. If you add 3 and 7 you get 10, not?) There is also the “becquerel,” Bq, equal to 1 decay per second, defined but almost never used. Why not “dpm,” disintegrations per minute, come to think of it? Why not indeed. The minute is just the additional unit the SI system needs, and using an acronym is great for creating confusion.

Of course the activity only tells you the amount of decays, not how bad the generated radiation is for your health. The “exposure” is the ionization produced by the radiation in a given mass of air, in SI units of Coulomb per kg. Exposure is very important for all people made of air. Of course, a better unit than a blasted SI one is needed, so the “roentgen” or “röntgen” R is defined to 2.58 10 $\POW9,{-4}$ C/kg. Why not?

But health-wise you may be more interested in the “absorbed dose” or “total ionizing dose” or “TID.” That is the radiation energy absorbed per unit mass. That would be in J/kg or “gray,” Gy, in SI units, but people really use the “rad” which is one hundredth of a gray.

If an organ or tissue absorbs a given dose of radiation, it is likely to be a lot worse if all that radiation is concentrated near the surface than if it is spread out. The “quality factor”

or the somewhat differently defined “radiation weighting factor” $w_{\rm {R}}$ is designed to correct for that fact. X-rays, beta rays, and gamma rays have radiation weighting factors (quality factors) of 1, but energetic neutrons, alpha rays and heavier nuclei go up to 20. Higher quality means worse for your health. Of course you already guessed that.

The bad effects of the radiation on your health are taken to be approximately given by the “equivalent dose,” equal to the average absorbed dose of the organ or tissue times the radiation weighting factor. It is in SI units of J/kg, called the “sievert” Sv, but people really use the “rem,” equal to one hundredth of a sievert. Note that the units of dose and equivalent dose are equal; the name is just a way to indicate what quantity you are talking about. It works if you can remember all these names.

To get the “effective dose” for your complete body, the equivalent doses for the organs and tissues must still be multiplied by “tissue weighting factors and summed. The weighting factors add up to one when summed over all the parts of your body. The ICRP defines “dose equivalent” different from equivalent dose. Dose equivalent is used on an operational basis. The personal dose equivalent is defined as the product of the dose at a point at an appropriate depth in tissue, (usually below the point where the dosimeter is worn), times the quality factor (not the radiation weighting factor).