19.5 Activity and Half-Life

19.5 Activity and Half-Life

We have seen that whenever an $\alpha$ or $\beta$-particle is emitted from a radioactive element, it is transformed into some other element. This radioactive decay process is quite random and is not subjected to any symmetry. Fluctuation in decay curves is also an evidence of its random nature. This means that we cannot foretell about any particular atom as to when will it decay. It could decay immediately or it may remain unchanged for thousands of years. Thus, we cannot say anything about the life of any particular atom of a radioactive element. While we cannot predict when a single nucleus will decay but in a very large sample the decay follows a pattern or consistency when we draw the number of radioactive nuclei against time (Fig. 19.7).

(Fig. 19.7)

The activity of a radioactive substance is the number of particles it emits per second. Each emission brings a change in the nucleus and is called decay rate. Its SI unit is Becquerel ($\mathrm{Bq}$). $1\,\mathrm{Bq} = 1$ decay per second The count rate by a nuclear radiation detector is the number of counts recorded per second. It has been found directly proportional to the activity. As time goes, the activity of a source decreases in a consistent manner.

The time taken for the activity of a sample to decrease to half of any starting value is called half life. Besides getting the definition of half-life, we can deduce two other conclusions from this example. These are, firstly no radioactive element can completely decay. It is due to the reason that in any half-life period, only half of the nuclei decay and in this way, an infinite time is required for all the atoms to decay. Secondly, the number of atoms decaying in a particular period is proportional to the number of atoms present in the beginning of the period. If the number of atoms to start with is large then a large number of atoms will decay in this period and if the number of atoms present in the beginning is small then less atoms will decay. We can represent these results with an equation. If at any particular time the number of radioactive atoms be $N$, then in an interval $\Delta t$, the number of decaying atom, $\Delta N$ is proportional to the time interval $\Delta t$ and the initial number of atoms $N$, i.e.,

or

where "$\lambda$" is the constant of proportionality and is called decay constant. Eq. 19.9 shows that if the decay constant of any element is large, then in a particular interval more of its atoms will decay and if the constant "$\lambda$" is small, then in that very interval less number of atoms will decay. Thus, decay constant of any element is equal to the fraction of the decaying atoms per unit time. The unit of the decay constant is $\mathrm{s^{-1}}$. The negative sign in Eq. 19.10 indicates the decrease in the number of atoms $N$. From Eq. 19.9 we can find activity:

The decay ability of any radioactive element is shown by the graphical curve (Fig. 19.7). We know that every radioactive element decays at a particular rate with time. If we draw a graph between number of atoms in the sample of the radioactive element present at different times and the time then a curve as shown in Fig. 19.7 will be obtained. This graph is called decay curve. After a period of one half-life $N_{o} / 2$ number of atoms of this radioactive element are left behind. If we wait further for another half-life period, then half of the remaining $N_{o} / 2$ atoms decay, and $\tfrac{1}{2} \times N_{o} /2 = (\tfrac{1}{2})^{2} N_{o}$ atoms remain behind. After the expiry of further period of a half-life, half of the remaining $(\tfrac{1}{2})^{2} N_{o}$ atoms decay. The number of atoms that remain un-decayed is $\tfrac{1}{2} \times (\tfrac{1}{2})^{2} N_{o} = (\tfrac{1}{2})^{3} N_{o}$. We can conclude from this example that if we have $N_{0}$ number of any radioactive element, then after a period of $n$ half-lives, the number of atoms left behind is $(\tfrac{1}{2})^{n} N_{o}$. In a similar way, we can use the general formula for decayed radioactive nuclides: After first half-life $= N_{0} - N_{0}/2 = N_{0}/2$ After second half-life $= N_{0} - N_{0}/4 = 3N_{0}/4$ After third half-life $= N_{0} - N_{0}/8 = 7N_{0}/8$ After $n$ half-lives;

It has been found that the estimate of decay of every radioactive element is according to the graph of Fig. 19.7, but the half-life of every radioactive element is different. For example, the half-life of uranium-$238$ is $4.5 \times 10^{9}$ years while the half-life of radium-$226$ is $1620$ years. The half-life of some radioactive elements is very small, for example, the half-life of radon gas is $3.8$ days and that of uranium-$239$ is $23.5$ minutes. From the above discussion, it is found that the estimate of any radioactive element can be made from its half-life or by determining its decay constant '$\lambda$'. It can be proved with the help of calculus that the following relations exist between the decay constant '$\lambda$' and the number of existing number of nucleons at any time $t$.

Where $N_{0}$ is the initial number of nuclides and $N$ is the undecayed nuclei after time '$t$'. When $N$ reduces to $N_{o}/2$, $t$ is known as half-life $T_{1/2}$. The solution of Eq. 19.12 gives,

Thus, if the decay constant '$\lambda$' of any radioactive element is known, its half-life can be found. Any stable element; besides the naturally occurring radioactive element, can be made radioactive. For this, very high energy particles are bombarded on the stable element. This bombardment excites the nuclei and the nuclei after becoming unstable become radioactive element. Such radioactive elements are called artificial radioactive elements.

For your information

lightbulb

Artificial Sun has been designed to mimic real Sun. It is an experimental advanced superconducting Tokamak (EAST) project in China. It uses powerful magnetic field to confine super heated plasma reaching temperature over million $^{\circ}\mathrm{C}$ in an attempt to fuse hydrogen isotopes into helium nucleus releasing clean energy, just like our Sun.

Example 19.2

The half-life of polonium ($_{84}^{209}\mathrm{Po}$) is $140$ days. If there are $1000$ atoms of polonium initially present in a sample, how many of them will decay in $280$ days.

Solution

Original atoms of polonium $= 1000$ Half-life of polonium $= T_{1/2} = 140$ days Number of atoms decaying in $140$ days $= 1000/2$ Number of polonium atoms left $= 1000 – 500 = 500$ Number of atoms decaying in next $140$ days $= 500/2 = 250$ Total atoms of polonium decaying in $280$ days $= 500 + 250 = 750$

Example 19.3

Americium-$241$ is an artificially produced radioactive element that emit $5.9\times 10^{5}$ g particles. Its sample of mass $5.1\,\mu\mathrm{g}$ is found to have an activity $5.9 \times 10^{5}\,\mathrm{Bq}$. Determine, for this sample: (i) the total sample of nuclei (ii) the decay constant (iii) half-life in years

Solution

(i) Using Avagadro's number $N_{A} = 6.02 \times 10^{23}$ Number of nucleons $= (\text{mass} \times N_{A}) / (\text{Number of nucleons})$ $= (5.1 \times 10^{-6}\,\mathrm{g} \times 6.02 \times 10^{-23}) / 241$ $= 1.27 \times 10^{16}$ (ii) Using Eq. 19.16 $A = -\lambda N$ Putting the values $5.9 \times 10^{5}\,\mathrm{s^{-1}} = -\lambda \times 1.27 \times 10^{16}$ $\lambda = 4.65 \times 10^{-11}\,\mathrm{s^{-1}}$ (iii) Half-life $T_{1/2} = 0.693 / \lambda$ $= 0.693 / (60 \times 60 \times 24 \times 365) \times 1.49 \times 10^{10}\,\mathrm{s}$ $= 1.49 \times 10^{10}\,\mathrm{s} / (60 \times 60 \times 24 \times 365)$ $= 472$ years