13.6 Radioactivity

A. H. Becquerel discovered radioactivity in 1896 purely by accident. While studying the fluorescence and phosphorescence of compounds irradiated with visible light, Becquerel observed an interesting phenomenon. After illuminating some pieces of uranium-potassium sulphate with visible light, he wrapped them in black paper and separated the package from a photographic plate by a piece of silver. When, after several hours of exposure, the photographic plate was developed, it showed blackening due to something that must have been emitted by the compound and was able to penetrate both black paper and the silver.

Experiments performed subsequently showed that radioactivity was a nuclear phenomenon in which an unstable nucleus undergoes a decay. This is referred to as radioactive decay. Three types of radioactive decay occur in nature :

$α$ -decay in which a helium nucleus $^{4}$ He is emitted;
$β$ -decay in which electrons or positrons (particles with the same mass as electrons, but with a charge exactly opposite to that of electron) are emitted;
$γ$ -decay in which high energy (hundreds of keV or more) photons are emitted.

Each of these decay will be considered in subsequent sub-sections.

13.6.1 Law of radioactive decay

In any radioactive sample, which undergoes α , β or γ -decay, it is found that the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample. If N is the number of nuclei in the sample and ∆ N undergo decay in time ∆ t then

$\begin{aligned} \frac{Δ N}{Δ t} \propto N \\ or, Δ N / Δ t = λ N, \end{aligned}$

where λ is called the radioactive decay constant or disintegration constant.

The change in the number of nuclei in the sample $^{*}$ is $d N = - Δ N$ in time $Δ t$ . Thus the rate of change of $N$ is (in the limit $Δ t \to 0$ ) $\frac{d N}{d t} = - λ N$ $or, \frac{d N}{N} = - λ d t$

Now, integrating both sides of the above equation,we get,

$\int_{N_{0}}^{N} \frac{d N}{N} = - λ \int_{t_{0}}^{t} d t$ $or, \ln N - \ln N_{0} = - λ (t - t_{0})$

Here $N_{0}$ is the number of radioactive nuclei in the sample at some arbitrary time $t_{0}$ and $N$ is the number of radioactive nuclei at any subsequent time $t$ . Setting $t_{0} = 0$ and rearranging Eq. (13.12) gives us

$\ln \frac{N}{N_{0}} = - λ t$ $N (t) = N_{0} e^{- λ t}$

Note, for example, the light bulbs follow no such exponential decay law. If we test 1000 bulbs for their life (time span before they burn out or fuse), we expect that they will ‘decay’ (that is, burn out) at more or less the same time. The decay of radionuclides follows quite a different law, the law of radioactive decay represented by Eq. (13.14).

The total decay rate $R$ of a sample is the number of nuclei disintegrating per unit time. Suppose in a time interval dt, the decay count measured is $Δ N$ . Then $d N = - Δ N$ .

The positive quantity R is then defined as $R = - \frac{d N}{d t}$ Differentiating Eq. (13.14), we get $R = λ N_{0} e^{- λ t}$

$R = R_{0} e^{- λ t}$

This is equivalant to the law of radioactivity decay, since you can integrate Eq. (13.15) to get back Eq. (13.14). Clearly, $R_{0} = λ N_{0}$ is the decay rate at $t = 0 .$ The decay rate $R$ at a certain time $t$ and the number of undecayed nuclei $N$ at the same time are related by

$R = λ N$

The decay rate of a sample, rather than the number of radioactive nuclei, is a more direct experimentally measurable quantity and is given a specific name: activity. The SI unit for activity is becquerel, named after the discoverer of radioactivity, Henry Becquerel.

1 becquerel is simply equal to 1 disintegration or decay per second. There is also another unit named “curie” that is widely used and is related to the SI unit as:

$\begin{aligned} 1 curie = 1 Ci = 3.7 \times 10^{10} decays per second \\ = 3.7 \times 10^{10} Bq \end{aligned}$

Different radionuclides differ greatly in their rate of decay. A common way to characterize this feature is through the notion of half-life. Half-life of a radionuclide (denoted by $T_{1 / 2}$ ) is the time it takes for a sample that has initially, say $N_{0}$ radionuclei to reduce to $N_{0} / 2$ . Putting $N = N_{0} / 2$ and $t = T_{1 / 2}$ in Eq. $(13.14),$ we get

$T_{1 / 2} = \frac{\ln 2}{λ} = \frac{0.693}{λ}$

Clearly if $N_{0}$ reduces to half its value in time $T_{1 / 2}$ , $R_{0}$ will also reduce to half its value in the same time according to Eq. (13.16).

Another related measure is the average or mean life $τ$ . This again can be obtained from Eq. (13.14). The number of nuclei which decay in the time interval $t$ to $t +$ $Δ t$ is $R (t) Δ t (= λ N_{0} e^{- λ t} Δ t) .$ Each of them has lived for time t. Thus the total life of all these nuclei would be $t λ N_{0} e^{- λ t}$ $Δ t .$ It is clear that some nuclei may live for a short time while others may live longer. Therefore to obtain the mean life, we have to sum (or integrate) this expression over all times from 0 to $\infty$ , and divide by the total number $N_{0}$ of nuclei at $t = 0 .$ Thus, $τ = \frac{λ N_{0} \int_{0}^{\infty} t e^{- λ t} d t}{N_{0}} = λ \int_{0}^{\infty} t e^{- λ t} d t$ One can show by performing this integral that $τ = 1 / λ$ We summarise these results with the following:

$T_{1 / 2} = \frac{\ln 2}{λ} = τ \ln 2$

Radioactive elements (e.g., tritium, plutonium) which are short-lived i.e., have half-lives much less than the age of the universe ( ∼ 15 billion years) have obviously decayed long ago and are not found in nature. They can, however, be produced artificially in nuclear reactions

13.6.2 Alpha decay

A well-known example of alpha decay is the decay of uranium $_{92}^{238} U$ to thorium $_{90}^{234}$ Th with the emission of a helium nucleus $_{2}^{4}$ He

$_{92}^{238} U \to_{90}^{234} Th +_{2}^{4} He (α - decay)$

In $α$ -decay, the mass number of the product nucleus (daughter nucleus) is four less than that of the decaying nucleus (parent nucleus), while the atomic number decreases by two. In general, $α$ -decay of a parent nucleus $_{z}^{A} X$ results in a daughter nucleus $_{z - 2}^{A - 4} Y$

$_{Z}^{A} X \to_{Z - 2}^{A - 4} Y +_{2}^{4} He$

From Einstein's mass-energy equivalance relation [Eq. (13.6)] and energy conservation, it is clear that this spontaneous decay is possible only when the total mass of the decay products is less than the mass of the initial nucleus. This difference in mass appears as kinetic energy of the products. By referring to a table of nuclear masses, one can check that the total mass of $_{90}^{234} Th$ and $_{2}^{4}$ He is indeed less than that of $_{92}^{238} U$ .

The disintegration energy or the Q-value of a nuclear reaction is the difference between the initial mass energy and the total mass energy of the decay products. For α-decay

$Q = (m_{X} - m_{Y} - m_{He}) c^{2}$

Q is also the net kinetic energy gained in the process or, if the initial nucleus X is at rest, the kinetic energy of the products. Clearly, Q> 0 for exothermic processes such as α-decay.

13.6.3 Beta decay

In beta decay, a nucleus spontaneously emits an electron $(β^{-}$ decay) or a positron $(β^{+}$ decay $)$ . A common example of $β^{-}$ decay is $_{15}^{32} P \to_{16}^{32} S + e^{-} + \bar{v}$ and that of $β^{+}$ decay is $_{11}^{22} Na \to_{10}^{22} Ne + e^{+} + v$

The decays are governed by the Eqs. (13.14) and $(13.15),$ so that one can never predict which nucleus will undergo decay, but one can characterize the decay by a half-life $T_{1 / 2} .$ For example, $T_{1 / 2}$ for the decays above is respectively $14.3 d$ and $2.6 y .$ The emission of electron in $β^{-}$ decay is accompanied by the emission of an antineutrino $(\bar{v});$ in $β^{+}$ decay, instead, a neutrino (v) is generated. Neutrinos are neutral particles with very small (possiblly, even zero) mass compared to electrons. They have only weak interaction with other particles. They are, therefore, very difficult to detect, since they can penetrate large quantity of matter (even earth) without any interaction.

In both $β^{-}$ and $β^{+}$ decay, the mass number A remains unchanged. In $β^{-}$ decay, the atomic number $Z$ of the nucleus goes up by $1,$ while in $β^{+}$ decay $Z$ goes down by $1 .$ The basic nuclear process underlying $β^{-}$ decay is the conversion of neutron to proton $n \to p + e^{-} + \bar{v}$ \text { while for } \beta^{+} \text {decay, it is the conversion of proton into neutron } $p \to n + e^{+} + v$ Note that while a free neutron decays to proton, the decay of proton to neutron [Eq. (13.25)] is possible only inside the nucleus, since proton has smaller mass than neutron.

13.6.4 Gamma decay

Like an atom, a nucleus also has discrete energy levels - the ground state and excited states. The scale of energy is, however, very different. Atomic energy level spacings are of the order of eV, while the difference in nuclear energy levels is of the order of MeV. When a nucleus in an excited state spontaneously decays to its ground state (or to a lower energy state), a photon is emitted with energy equal to the difference in the two energy levels of the nucleus. This is the so-called gamma decay. The energy (MeV) corresponds to radiation of extremely short wavelength, shorter than the hard X-ray region.

Typically, a gamma ray is emitted when a $α$ or $β$ decay results in a daughter nucleus in an excited state. This then returns to the ground state by a single photon transition or successive transitions involving more than one photon. A familiar example is the successive emmission of gamma rays of energies $1.17 MeV$ and $1.33 MeV$ from the deexcitation of $_{28}^{60} Ni$ nuclei formed from $β^{-}$ decay of $_{27}^{60} Co$