Every morning a nuclear medicine department in a large hospital, in Delhi or in Chennai, prepares doses of technetium-99m. A small amount injected into a patient gives out gamma rays. A camera outside the body turns them into a picture of the bones, the heart or the kidneys. The dose must be measured just before the injection. A dose that was right at 8 in the morning is too weak by 2 in the afternoon. The vial has not even been opened.
Chance, one nucleus at a time
Watch a single radioactive nucleus. It may decay in the next second, or it may last a thousand years. Nothing in the nucleus counts down. A nucleus that has already survived for an hour is no nearer to its end than a fresh one. It has no age. Its chance of decaying in the next second is fixed by what kind of nucleus it is, and by nothing else.
That sounds as if nothing could be predicted. But a sample holds an enormous number of nuclei, and each one gives the same fixed chance. Toss one coin and you cannot say how it will land. Toss a million coins and very close to half will be heads. In the same way, a large sample of nuclei decays in a smooth and exactly predictable way.
Play Fig 3.1 and try to guess which dot goes next; there is no pattern to find. Yet the staircase of the count hugs the smooth curve. With 400 nuclei the count wanders by about ten either side of the curve. A real sample has around \(10^{13}\) nuclei or more, and there the wandering is far too small to measure.
Two nuclei of the same radioactive kind are watched. Nucleus P was formed a year ago, and nucleus Q was formed a minute ago. Which is more likely to decay in the next second?
The decay constant
Call the fixed chance of decay per unit time the decay constant \(\lambda\). In a short time \(dt\), one nucleus decays with probability \(\lambda\,dt\). Among \(N\) nuclei, about \(\lambda N\,dt\) decay in that time, and the number left falls by the same amount. The rate at which \(N\) falls is therefore proportional to \(N\) itself.
The number of nuclei that decay per unit time is proportional to the number of undecayed nuclei present: \(dN/dt = -\lambda N\). The constant \(\lambda\) is the decay constant of the nuclide.
The unit of \(\lambda\) is one over time: per second, per hour or per year. A value of \(\lambda = 0.1\) per hour means that each nucleus has about a 10 % chance of decaying within the next hour. Twice as many nuclei give twice as many decays each hour, and a nucleus with a larger \(\lambda\) decays sooner on average.
The chance of decay in a time \(t\) equals \(\lambda t\) only when \(\lambda t\) is small. With \(\lambda = 0.1\) per hour, the chance in one hour is \(1 - e^{-0.1} = 9.5\,\%\), close to but not exactly 10 %. Over ten hours it is not 100 %, but \(1 - e^{-1} = 63\,\%\). A chance can never exceed one.
From the rate to the number left
The law gives the rate of change of \(N\). To find \(N\) itself at any time, separate the variables and integrate, as in the derivation below. At \(t = 0\) there are \(N_0\) nuclei.
\[ \dfrac{dN}{N} = -\lambda\,dt \]
Why. divide the law \(dN/dt = -\lambda N\) by \(N\) and multiply by \(dt\), so each side holds one variable.
\[ \int_{N_0}^{N} \dfrac{dN}{N} = -\lambda \int_0^t dt \]
Why. integrate from the start, \(N_0\) at time 0, to \(N\) at time \(t\).
\[ \ln\dfrac{N}{N_0} = -\lambda t \]
Why. the integral of \(1/N\) is \(\ln N\), and \(\ln N - \ln N_0 = \ln(N/N_0)\).
\[ N = N_0 e^{-\lambda t} \]
Read it. take the exponential of both sides.
The exponential has two properties worth remembering. First, in any interval of the same length, \(N\) falls by the same fraction. If a sample falls to one third in 10 minutes, it falls to one ninth in 20 minutes, whatever it started with. Second, \(N\) never quite reaches zero. Each interval removes a fraction of what is left, so something always remains, until the last few nuclei decay one by one.
The second form of the law is useful in the laboratory. Taking the logarithm turns the curve into a straight line: \(\ln N = \ln N_0 - \lambda t\). Plot \(\ln N\) against \(t\), and the points fall on a line whose slope is \(-\lambda\) and whose intercept is \(\ln N_0\). A straight line is far easier to test and to read than a curve.
For a sample, \(\ln N\) is plotted against \(t\) in hours. The points lie on a straight line from \(\ln N = 18.0\) at \(t = 0\) to \(\ln N = 16.5\) at \(t = 5\,\text{h}\). Find \(\lambda\) and \(N_0\).
The line is \(\ln N = \ln N_0 - \lambda t\). Its slope is \(-\lambda\), and its intercept on the \(\ln N\) axis is \(\ln N_0\).
Slope \(= (16.5 - 18.0)/(5 - 0) = -0.3\) per hour.
So \(\lambda = 0.3\) per hour. The intercept is \(\ln N_0 = 18.0\), so \(N_0 = e^{18} = 6.6 \times 10^7\).
\(\lambda = 0.3\) per hour and \(N_0 = 6.6 \times 10^7\) nuclei. Values used: the two points of the line, \(e^{18} = 6.57 \times 10^7\).
A nuclide has \(\lambda = 0.05\) per second. What fraction of a sample is left undecayed after \(20\,\text{s}\)? Take \(e^{-1} = 0.368\).
Activity: decays per second
A counter cannot see how many nuclei are left. It counts decays. The number of decays per second is the activity of the sample, \(A = -dN/dt\). From the decay law, \(A = \lambda N\). The activity is proportional to the number left, so it falls in exactly the same way: \[ A = \lambda N_0\,e^{-\lambda t} = A_0\,e^{-\lambda t} \]
The SI unit of activity is the becquerel, Bq, one decay per second. A becquerel is tiny: the potassium in a human body gives a few thousand becquerel. The older unit, the curie, is \(3.7 \times 10^{10}\,\text{Bq}\). It was chosen as about the activity of one gram of radium, as the example shows.
In the figure above, the activity starts at \(\lambda N_0 = 0.3465 \times 8000 = 2772\) decays per hour. After \(2\,\text{h}\), \(N = 4000\) and the activity is \(1386\) per hour, half as much. The steepness of the curve and the number of nuclei fall together, because the one is \(\lambda\) times the other.
Radium-226 has a half-life of 1600 years, and its decay constant is \(\lambda = 0.693/T_{1/2}\) (the next lesson shows why). Find the activity of \(1\,\text{g}\) of radium-226. Take \(1\,\text{y} = 3.156 \times 10^7\,\text{s}\) and \(N_A = 6.02 \times 10^{23}\) per mole.
Find \(N\) from the mass and \(\lambda\) from the half-life, in seconds. Then \(A = \lambda N\).
\(N = (1/226) \times 6.02 \times 10^{23}\), and \(\lambda = 0.693/(1600 \times 3.156 \times 10^7\,\text{s})\).
\(N = 2.66 \times 10^{21}\) and \(\lambda = 1.37 \times 10^{-11}\) per second. So \(A = 1.37 \times 10^{-11} \times 2.66 \times 10^{21} = 3.66 \times 10^{10}\,\text{Bq}\).
One gram of radium gives about \(3.7 \times 10^{10}\) decays per second, one curie. Values used: \(\ln 2 = 0.693\), \(1\,\text{y} = 3.156 \times 10^7\,\text{s}\), \(N_A = 6.02 \times 10^{23}\) per mole.
A sample holds \(8 \times 10^{20}\) nuclei of a nuclide whose decay constant is \(1.6 \times 10^{-6}\) per second. Find its activity, in units of \(10^{15}\,\text{Bq}\).
Samples P and Q hold the same number of nuclei. The decay constant of P is twice that of Q. Compare their activities now.
Now return to the hospital. Technetium-99m has \(\lambda = 0.1155\) per hour. A dose of \(800\,\text{MBq}\) measured at 8 in the morning has, six hours later at 2 in the afternoon, an activity of \(800\,e^{-0.1155 \times 6} = 800\,e^{-0.693} = 400\,\text{MBq}\). Half of it has gone, with the vial still sealed. A patient booked for the afternoon needs twice the morning's amount of technetium.
A counter near a radioactive sample records 4000 counts per minute today. Why will it record fewer tomorrow, even if the sample is not touched?
The count rate follows the activity, and the activity is \(\lambda N\). Every decay removes one nucleus, so \(N\) falls with time, and with it the number of decays per minute. The chance of decay for each nucleus stays the same; there are simply fewer nuclei left to decay.
- Thinking an old nucleus is more likely to decay soon.A nucleus has no age. Its chance of decay per second is the same at every moment.
- Taking the chance of decay in a time \(t\) as \(\lambda t\) for any \(t\).That holds only when \(\lambda t\) is small. In general the chance is \(1 - e^{-\lambda t}\).
- Mixing units of time between \(\lambda\) and \(t\).\(\lambda t\) must be a pure number. Use \(\lambda\) per hour with \(t\) in hours, or per second with \(t\) in seconds.
- Confusing the activity with the number of nuclei.Activity is decays per second, \(A = \lambda N\). Find \(N\) from \(A\) by dividing by \(\lambda\).