Half-life and radiation protection principles

Atomic nuclei of the same element with different numbers of neutrons are isotopes; some of them are unstable and decay spontaneously, emitting alpha or beta particles and gamma rays.

Decay of an individual nucleus is a random event, but for a macroscopic sample the number of decays per second is proportional to the number of nuclei:

$$\frac{dN}{dt}=-\lambda N,\qquad N(t)=N_0 e^{-\lambda t}.$$

The half-life $T_{1/2}$ is the time after which exactly half of the nuclei remain:

$$T_{1/2}=\frac{\ln 2}{\lambda},\qquad N(t)=N_0 2^{-t/T_{1/2}}.$$

Activity $A=\lambda N$ is measured in becquerels (Bq), while biological effects are described by absorbed dose in grays (Gy = J/kg) and effective dose in sieverts (Sv).

Radiation protection rests on three levers — time, distance, and shielding:

$$I(r)=\frac{I_0 r_0^2}{r^2},\qquad I(x)=I_0 e^{-\mu x},$$

and the ALARA principle requires keeping the dose as low as reasonably achievable.

In alpha decay, momentum conservation (the same idea as in collisions) splits the energy $Q$ between the alpha particle and the recoil nucleus in inverse proportion to their masses:

$$E_\alpha=Q\frac{M_2}{M_2+m_\alpha},$$

where $M_2$ is the daughter-nucleus mass.

A sample contains $N_0=6.0\times10^{20}$ nuclei of an isotope with half-life $T_{1/2}=8.0\ \mathrm{d}$. How many nuclei remain after 24 days, and what is the decay constant $\lambda$?

A point gamma source produces an equivalent dose rate of $40\ \mathrm{\mu Sv/h}$ at a distance of $2.0\ \mathrm{m}$.

(a) What is the dose rate at $5.0\ \mathrm{m}$?

(b) How many layers of lead with half-value layer $d_{1/2}=1.2\ \mathrm{cm}$ are needed to reduce the dose rate below $1.0\ \mathrm{\mu Sv/h}$?

A $^{226}\mathrm{Ra}$ nucleus undergoes alpha decay: $^{226}\mathrm{Ra}\to{}^{222}\mathrm{Rn}+\alpha$ with total decay energy $Q=4.87\ \mathrm{MeV}$. Assuming the parent $^{226}\mathrm{Ra}$ nucleus is at rest before the decay, find the kinetic energies of the alpha particle and the recoil nucleus. Explain why this alpha emitter is dangerous when inhaled but easily shielded from outside.

Safe self-study: do not use real radioactive sources.

1. Open the PhET simulation “Alpha Decay” (or a similar radioactive-decay model) and start it with a large number of nuclei (for example, 100).
2. Record the number of undecayed nuclei $N$ at equal time intervals until roughly a quarter of the initial amount remains.
3. Plot $\ln N$ against time: the points should form a straight line with slope $-\lambda$; find $T_{1/2}=\ln 2/\lambda$ and compare it with the value shown by the simulation.
4. In the same simulation, place different materials (paper, aluminium, lead) and different thicknesses between the source and the detector, then double the distance. Record what reduces the count rate more strongly — distance or shielding.
5. State three practical ALARA rules for an imaginary laboratory: less time near the source, greater distance, and the right shield.

Show that exponential decay becomes a straight line on a semi-log plot, and that the dose rate from a point source falls as $1/r^2$ while a shield attenuates it in halving steps.

Two-panel diagram.

• Panel (a): a graph with $\ln N$ vertical and time $t$ horizontal; a straight line with slope $-\lambda$; the time axis marks $T_{1/2}$, $2T_{1/2}$, $3T_{1/2}$, and the $\ln N$ axis marks $\ln N_0$, $\ln N_0-\ln 2$, and so on.
• Panel (b): a sketch of a point source at the centre, two circles of radii $r$ and $2r$ with the same number of rays passing through them; beside it, bars for dose rates $I$ and $I/4$, and on the right a shield plate labelled $d_{1/2}$ with decreasing arrows $I\to I/2\to I/4$.

A banana contains about 450 mg of potassium. The fraction of the radioactive isotope $^{40}\text{K}$ in natural potassium is 0.0117 %, and its half-life is $T_{1/2}=1.25\times 10^9$ years. The activity of such a banana is about 15 Bq, that is, 15 decays per second. Eating one banana gives a person an effective dose of about 0.1 µSv, whereas the natural background gives about 2.4 mSv per year. Because potassium in the body is regulated, the activity of $^{40}\text{K}$ in the body does not fall with time according to this half-life.

Here the relation between activity and the number of nuclei is used: $A=\lambda N$, where $\lambda=\frac{\ln 2}{T_{1/2}}$. The huge half-life $1.25\times 10^9$ years gives a tiny decay constant $\lambda\approx 1.76\times 10^{-17}\ \text{s}^{-1}$, so over a human lifetime the number of nuclei $N$ is practically unchanged. The dose of 0.1 µSv is compared with the annual background of 2.4 mSv, which demonstrates the principle of estimating radiation hazard through dose.

In a basement with a volume of $60\ \text{m}^3$, the volumetric activity of radon $^{222}\text{Rn}$ is $200\ \text{Bq/m}^3$. The total activity of radon in the basement is $A=200\ \text{Bq/m}^3\times 60\ \text{m}^3=12\,000\ \text{Bq}$. The half-life of radon is $T_{1/2}=3.82$ days. If the basement is closed, after 3.82 days half of the radon nuclei remain, but radon continuously enters from the soil. Opening a window for 30 minutes can reduce the concentration several times due to ventilation, not only due to decay.

The half-life sets the rate at which the number of nuclei decreases: $N(t)=N_0 2^{-t/T_{1/2}}$. At $t=3.82$ days, $N=N_0/2$. However, in a real room the decrease of radon is also determined by ventilation: the protection principle “time — distance — shielding” is applied here as reducing the time spent and removing the source (ventilation). The short half-life of radon means that it quickly turns into daughter isotopes (for example, $^{218}\text{Po}$), which give the main dose upon inhalation.

To inspect a bridge weld, a source of $^{192}\text{Ir}$ with an initial activity $A_0=3.7\ \text{TBq}$ ($100\ \text{Ci}$) is used. Its half-life is $T_{1/2}=73.8$ days. After 74 days the activity becomes $A=1.85\ \text{TBq}$, so the exposure time for the same image density must be doubled. At a distance $r=1$ m the dose rate from the source is about $10\ \text{mSv/h}$; at a distance $r=20$ m it drops by a factor of $20^2=400$ and becomes $25\ \text{µSv/h}$. In 8 minutes at such a distance the dose will be about $25\ \text{µSv/h}\times \frac{8}{60}\ \text{h}\approx 3.3\ \text{µSv}$. The half-value layer of lead for the gamma radiation of $^{192}\text{Ir}$ is about 12 mm.

The half-life determines the activity at any moment: $A(t)=A_0 2^{-t/T_{1/2}}$. At $t=74$ days, $t/T_{1/2}\approx 1.00$, hence $A\approx A_0/2$. Protection is based on three principles: time (reduce exposure), distance (inverse-square law $I\propto 1/r^2$), and shielding (half-value layer). The calculation $10/20^2=0.025\ \text{mSv/h}=25\ \text{µSv/h}$ shows how distance replaces a heavy lead shield.