Two ways out
Some nuclei can decay in two different ways. Bismuth-212 has a half-life of 61 minutes. It gives out a beta particle in 64 % of its decays and an alpha particle in the other 36 %. Each route has its own chance per unit time, \(\lambda_1\) for one and \(\lambda_2\) for the other. In a short time \(dt\), a nucleus decays by the first route with chance \(\lambda_1 dt\), or by the second with chance \(\lambda_2 dt\).
A sample of \(N\) nuclei loses \(\lambda_1 N\) per second by the first route and \(\lambda_2 N\) by the second. So \[ \dfrac{dN}{dt} = -\lambda_1 N - \lambda_2 N = -(\lambda_1 + \lambda_2)N \] This is the ordinary decay law with the decay constant \(\lambda = \lambda_1 + \lambda_2\). The nucleus decays exponentially, faster than either route alone would make it.
Divide each \(\lambda\) into \(\ln 2\) and the half-lives follow: \(1/T_{1/2} = 1/T_1 + 1/T_2\). Half-lives combine like resistors in parallel, and the result is shorter than either. Routes that alone would give \(3\,\text{h}\) and \(6\,\text{h}\) together give \(1/(1/3 + 1/6) = 2\,\text{h}\). The mean lives \(\tau = 1/\lambda\) combine in the same way.
Every nucleus decays exactly once, by one route or the other. The fraction that takes route 1 is its share of the total chance, \(\lambda_1/(\lambda_1 + \lambda_2)\). This fraction is the branching ratio of route 1. The faster route takes the larger share.
A nucleus decays by alpha emission and by beta emission. Alone, the alpha route would give a half-life of \(4\,\text{h}\), and the beta route one of \(12\,\text{h}\). Find the half-life of the nucleus and the fraction of its decays that are alpha decays. A sample starts with \(8 \times 10^{18}\) nuclei; how many alpha decays happen in the first \(6\,\text{h}\)?
The decay constants add: \(\lambda = \lambda_\alpha + \lambda_\beta\), with each \(\lambda = 0.693/T\).
\(1/T_{1/2} = 1/4 + 1/12 = 4/12\) per hour. The alpha fraction is \(\lambda_\alpha/\lambda = (1/4)/(1/3)\).
\(T_{1/2} = 3\,\text{h}\), and the alpha fraction is \(3/4\). In \(6\,\text{h}\), two half-lives, \(\tfrac34\) of the nuclei decay: \(6 \times 10^{18}\). Three quarters of these, \(4.5 \times 10^{18}\), are alpha decays.
The half-life is \(3\,\text{h}\), 75 % of the decays are alpha decays, and \(4.5 \times 10^{18}\) alpha decays happen in the first 6 hours. Values used: \(\lambda = 0.693/T\); \(\ln 2\) cancels.
A nucleus X decays to Y by one route and to Z by another. Alone, the route to Y would give a half-life of \(2\,\text{h}\), and the route to Z one of \(6\,\text{h}\). Find the half-life of X, in minutes.
The mean lives of a nucleus for its two decay routes, taken alone, are \(3\,\text{h}\) and \(6\,\text{h}\). Find its mean life, in hours.
Made and lost at the same time
In a reactor, a target placed among the neutrons turns into a radioactive nuclide at a steady rate \(R\), in nuclei per second. As soon as some are made, they start to decay, at the rate \(\lambda N\). The number present changes at the difference of the two rates: \[ \dfrac{dN}{dt} = R - \lambda N \]
At first \(N\) is small, the loss \(\lambda N\) is small, and \(N\) grows almost at the full rate \(R\). As \(N\) grows, so does the loss. The growth slows, and it stops when the loss equals the making, \(\lambda N = R\). The number then stays at \(R/\lambda\) for as long as production goes on.
\[ \dfrac{dN}{R - \lambda N} = dt \]
Why. separate the variables of \(dN/dt = R - \lambda N\).
\[ -\dfrac{1}{\lambda}\ln\dfrac{R - \lambda N}{R} = t \]
Why. integrate from \(N = 0\) at \(t = 0\); at the start \(R - \lambda N = R\).
\[ R - \lambda N = R\,e^{-\lambda t} \]
Why. multiply by \(-\lambda\) and take the exponential of both sides.
\[ N = \dfrac{R}{\lambda}\left(1 - e^{-\lambda t}\right) \]
Read it. solve for \(N\).
In each half-life, \(N\) closes half of the gap that remains to \(R/\lambda\). So after 1, 2, 3 and 4 half-lives it stands at 50, 75, 87.5 and 93.75 % of its final value. Running a reactor target for more than about four half-lives gains very little. At saturation the activity is \(\lambda N = R\): the sample decays exactly as fast as it is made.
A nuclide is made at a steady rate in a reactor, starting from none. What percentage of its final number is present after four half-lives?
Dhruva, the 100 MW research reactor of the Bhabha Atomic Research Centre at Trombay in Mumbai, makes radioactive isotopes for hospitals and industry. Targets are lowered into its flood of neutrons and left there while the new nuclide builds up. Suppose a target makes nuclei of half-life 8 days at a steady \(2 \times 10^6\) per second. The number would level off at \(R/\lambda = 2.0 \times 10^{12}\), with an activity of \(2 \times 10^6\,\text{Bq}\). After 24 days, three half-lives, the target already holds 87.5 % of that. Another week in the reactor would add little, so the target is taken out and processed.
A parent that feeds a daughter
Now let the producer itself be radioactive. A parent A decays to a daughter B, which is radioactive too and decays to a stable C. The parent decays undisturbed, \(N_A = N_0e^{-\lambda_A t}\). The daughter is made at the rate at which A decays and lost at its own rate of decay: \[ \dfrac{dN_B}{dt} = \lambda_A N_A - \lambda_B N_B \]
Starting with pure A, the daughter first grows, because at the start it is made and hardly lost. It reaches a peak at the moment its two rates balance, \(\lambda_A N_A = \lambda_B N_B\). After that the parent's supply has fallen below the daughter's decay, and \(N_B\) falls. The stable end C only gains, at the rate \(\lambda_B N_B\).
Secular equilibrium
Move the parent's half-life in the figure above up to \(20\,\text{h}\) and the daughter's down to \(0.5\,\text{h}\). The parent now hardly falls during the whole graph. After a few of the daughter's half-lives, the daughter settles into a fixed small fraction of the parent. The readout shows \(N_B/N_A = 0.026\) at \(12\,\text{h}\), close to \(T_B/T_A = 0.025\). The daughter now decays exactly as fast as it is made.
This is secular equilibrium. The two activities are equal, \(\lambda_A N_A = \lambda_B N_B\), and so the numbers are in the ratio of the half-lives: \[ \dfrac{N_B}{N_A} = \dfrac{\lambda_A}{\lambda_B} = \dfrac{T_B}{T_A} \] In the uranium-238 series, thorium-234 has a half-life of 24.1 days. In an old uranium ore there is one thorium-234 nucleus for every \(4.47 \times 10^9 \times 365.25/24.1 = 6.8 \times 10^{10}\) uranium-238 nuclei, yet the two give equal activities.
A sealed capsule holds radium-226 (half-life 1600 years) with an activity of \(1.85 \times 10^{10}\,\text{Bq}\). Its daughter, radon-222 (half-life 3.8 days), cannot escape. After a month, find the activity of the radon, the number of radon atoms and their mass. \(6.02 \times 10^{23}\) atoms of radon have a mass of \(222\,\text{g}\).
A month is nearly eight radon half-lives, so the radon has reached secular equilibrium with the radium.
Equal activities: \(A_{Rn} = A_{Ra}\). Then \(N_{Rn} = A_{Rn}/\lambda_{Rn}\), with \(\lambda_{Rn} = 0.693/(3.8 \times 86400\,\text{s})\).
\(A_{Rn} = 1.85 \times 10^{10}\,\text{Bq}\). \(N_{Rn} = 1.85 \times 10^{10} \times 3.8 \times 86400/0.693 = 8.8 \times 10^{15}\). Mass \(= 8.8 \times 10^{15} \times 222/(6.02 \times 10^{23}) = 3.2 \times 10^{-6}\,\text{g}\).
The radon has the full \(1.85 \times 10^{10}\,\text{Bq}\) of the radium, but only \(3.2\,\mu\text{g}\) of it is present. Values used: \(\ln 2 = 0.693\), \(1\,\text{day} = 86400\,\text{s}\), \(6.02 \times 10^{23}\) per mole.
A parent of half-life 5000 years feeds a daughter of half-life 10 days, and the pair is in secular equilibrium. Find \(N_B/N_A\), in units of \(10^{-6}\). Take 1 year as 365 days.
In a chain A → B → C starting from pure A, the number of B nuclei is at its greatest when
In the radium capsule of the example, three micrograms of radon give as many decays per second as half a gram of radium. Explain how so little can be so active.
Activity is \(\lambda N\), the number of nuclei times the chance of decay of each. The short-lived daughter has a huge \(\lambda\) and a tiny \(N\); the long-lived parent has a tiny \(\lambda\) and a huge \(N\). The products are equal. The daughter is like a small tank with a wide outlet fed by a slow tap: the level stays low, but the outflow equals the inflow.
Hospitals use a chain every day. Molybdenum-99, half-life 66 hours, decays to the technetium-99m of lesson 3, half-life 6 hours. Molybdenum made in a reactor reaches a hospital in a shielded column called a generator. Each morning the technetium is washed out, and the parent then breeds a fresh supply. By the formula above, the technetium is back at its peak about 23 hours later, in time for the next morning.
- Adding the half-lives of two decay routes.The decay constants add. The half-lives combine as \(1/T = 1/T_1 + 1/T_2\), which is shorter than either.
- Thinking the daughter's number is largest when parent and daughter are equal in number.The peak comes where the activities are equal, \(\lambda_A N_A = \lambda_B N_B\).
- Expecting secular equilibrium to mean equal numbers of nuclei.It means equal activities. The numbers are in the ratio of the half-lives.
- Leaving a target in a reactor for many half-lives to get more activity.The activity levels off at R. After about four half-lives little more is gained.