Radioactivity

RADIOACTIVITY

It was observed by Henri Becquerel in i896 that some minerals like pitchblende emit radiation spontaneously, and this radiation can blacken photographic plates if photographic plate is wrapped in light proof paper. He called this phenomenon radioactivity. It was observed that radioactivity was not affected by temperature, pressure or chemical combination, it could therefore be a property that was intrinsic to the atom-more specifically to its nucleus. Detailed studies of radioactivity resulted in the discovery that radiation emerging from radioactive substances were of three types : $\alpha$-rays, $\beta$ -rays and $\gamma$ -rays.

$\alpha$-rays were found to be positively charged, $\beta$-rays (mostly) negatively charged, and $\gamma$ - rays uncharged. Some of the properties of $\alpha ,\beta$ and $\gamma$– rays, are explain below:

(i) Alpha Particles: An $\alpha$-particle is a helium nucleus, i.e. a helium atom which has lost two electrons. It has a mass about four times that of a hydrogen atom and carries a charge +2e. The velocity of $\alpha$-particles ranges from 5 to 7 percent of the velocity of light. These have very little penetrating power into a material medium but have a very high ionising power.

(ii) Beta Particles: $\beta$-particles are electrons moving at high speeds. These have greater (compared to $\alpha$-particles) penetrating power but less ionising power. Their emission velocity is almost the velocity of light. Unlike $\alpha$-particles, they have a spectrum of energy, i.e., beta particles possess energy from a certain minimum to a certain maximum value.

(iii) Gamma Rays: $\gamma$-rays are electromagnetic waves of wavelength of the order of $\sim$10^–12 m. These have the maximum penetrating power (even more than X–ray) and the least ionising power. These are emitted due to transition of excited nucleus from higher energy state to lower energy state.

The other simple laws of radioactivity that were discovered by were the Rutherford-Soddy displacement laws:

(a) During a $\alpha$ decay, the daughter element was always two position below the parent in the periodic table; the mass number of the daughter nucleus was 4 units smaller than that of the parent nucleus. ${}_a^{b}X{=}_{a-2}^{b-4}Y+\alpha$

(b) During $\beta$-decay, the daughter nucleus had the same mass number as the parent while its atomic number was greater than that of the parent by 1 (smaller by 1 unit in ${\beta }^{+}$ decay). ${}_a^{b}X\to {}_{a+1}^{b}Y+{\beta }_{-}$

(c) $\gamma$ emission did not result in change of atomic number or mass number.

Radioactivity is observed to be a random process, it cannot be predicted when a particular nucleus will decay. One can only discuss the probability of its decay.

RADIOACTIVE DECAY LAW

The number of atoms disintegrating per second $\left(\frac{dN}{dt}\right)$ is directly proportional to the number of atoms (N) present at that instant. For most radioisotopes, $\lambda$. is very small in the SI system it take a large number N($\sim$ Avogadro number, 10²³) to get any significant activity.

$\frac{dN}{dt}=-\lambda N$ ….(i)

Where $\gamma$ is the radioactivity decay constant.

If N_o is the number of radioactive atoms present at a time t = 0, and N is the number at the end of time t, then

$N=N_0{e}^{-\lambda \mathbf{t}}$

The term $\left(-\frac{dN}{dt}\right)$is called the activity of a radioactive substance and is denoted by 'A'.

Units: 1 Becquerel (Bq) = 1 disintegration per second (dps)

1 curie (Ci) = 3.7 x 10¹⁰ dps

1 Rutherford =10⁶ dps

The changes in N, due to radioactivity, are very small in 1 s compared with the value of N 'its self.

Therefore. N may be approximated by means of a continuous function N(t).

For the case considered in equation (i), we can write,

A = rate of decrease of N = $\lambda N$

or $-\frac{dN}{dt}=\lambda N$ .....(ii)

Note the equation (ii) is valid only for large value of N, typically –10¹⁸ or so.

If the number of atoms at time t = 0 (initial) is N = N₀^, then one can solve equation (ii):

$\underset{N_0}{\overset{N}{\int }}\frac{dN}{N}=-\underset{0}{\overset{t}{\int }}\lambda dt$

or In $\frac{N}{N_0}=-\lambda t$

or N(t) = N₀ $e^{-\lambda t}$ ......(iii)

The activity A (t) = $\lambda N$(t) and is given by

A (t) = $\lambda N_0{e}^{-\lambda t}$

= A₀ $e^{-\lambda t}$ taking (A₀ = $\lambda$N₀) ....(iv)

The variation of N is represented below:

Graphs of A vs t or In A vs t are similar to the above graphs

HALF-LIFE

A significant feature of the above graph is the time in which the number of active nuclei (N) is halved. This is independent of the starring value, N = N₀. Let us compute this time:

$\frac{N_0}{2}=N_0{e}^{-\lambda r^{\prime }}$

$\therefore$ $e^{-\lambda t^{\prime }}$ = $\frac{1}{2}$

or, $\lambda t^{\prime }$ = In 2 = 0.693 (approx)

or, t' = $T_{1/2}$ = $\frac{0.693}{\lambda }$

This half-life represents the time in which the number of radioactive nuclei falls to $\frac{1}{2}$ of its starting value. Activity, being proportional to the number of active nuclei, also has the same half-life.

Illustration 1: A count rate-meter is used to measure the activity of a given sample. At one instant the meter shows 4750 counts per minute. Five minutes later it shows 2700 counts per minute. Find:

(a) decay constant

(b) the half life of the sample.

Solution: Initial activity = A₀ = dN/dt at t = 0

Final activity = A_t = dN/dt at t = t

${\frac{dN}{dt}|}_{t=0}=\lambda N_0\&{\frac{dN}{dt}|}_{t=5}=\lambda N_t$

$\Rightarrow \frac{4750}{2700}=\frac{N_0}{N_t}$

$\text{Using }\lambda t=2.303\log \frac{N_0}{N_t}$

$\Rightarrow \lambda \left(5\right)=2.303\log \frac{4750}{2700}\Rightarrow \frac{2.303}{5}\log \frac{4750}{2700}=0.1129{\min }^{-1}$

$\Rightarrow t_{1/2}=\frac{0.693}{0.1129}=6.14\min$

MEAN-LIFE

A very significant quantity that can be measured directly for small numbers of atoms is the mean lifetime. If there are n active nuclei, (atoms) (of the same type, of course), the mean life is

$\overline{\tau }=\frac{{\tau }_1+{\tau }_2+.......+{\tau }_2}{n}$

Where ${\tau }_1$, ${\tau }_2$,....... ${\tau }_n$ represent the observed lifetime of the individual nuclei and n is a very large number. It can also be calculated as a weighted average:

$\overline{\tau }=\frac{{\tau }_1{N}_1+{\tau }_2{N}_2+.......+{\tau }_n{N}_n}{N_1+.....+N_n}$

Where N₁ nuclei live for time ${\tau }_1$,

N₂ nuclei live for time ${\tau }_2,$....... and so on.

This quantity may be related with $\lambda$.

Using calculus (vi) may be rewritten as:

$\overline{\tau }=\frac{\int t\left|dN\right|}{\int \left|dN\right|}$

Where |dN| is the number of nuclei decaying between t, t + dt; the modulus sign is required to ensure that it is positive.

dN = $-\lambda N_0{e}^{-\lambda t}dt$

and |dN| = $\lambda N_0{e}^{-\lambda t}dt$

$\therefore$ $\overline{\tau }=\frac{\underset{0}{\overset{\infty }{\int }}t\lambda N_0{e}^{-\lambda t}dt}{\underset{0}{\overset{\infty }{\int }}\lambda N_0{e}^{-\lambda t}dt}$

= $\frac{\underset{0}{\overset{\infty }{\int }}t{e}^{-\lambda t}dt}{\underset{0}{\overset{\infty }{\int }}{e}^{-\lambda t}dt}$ = $\frac{\left(\frac{1}{{\lambda }^2}\right)}{\left(\frac{1}{\lambda }\right)}$ (integrating the numerator by parts)

or $\overline{\tau }=\frac{1}{\lambda }$

Illustration 2: Cascade Decays: Frequently, a nucleus A decays into another nucleus B, which again decays into C, and so on until the series ends in a stable nuclide. These are known as cascade decays. We will consider a simple example of a cascade decay where a nuclide a decays into B, and then, B decays into C:

$A\stackrel{{\lambda }_A}{\to }B\stackrel{{\lambda }_B}{\to }C$

Suppose that there are N₀ atoms of A to begin with the problem is to find the numbers 0f atoms of A, B, C respectively as a function of time t.

Solution: Suppose that, at time t, there are N_A atoms of A. N_b atoms of B. and We write the equations:

$\frac{d{N}_A}{dt}=$ –activity of A = $-{\lambda }_A{N}_A$ .....(i)

$\frac{d{N}_B}{dt}$ = + (activity of A) – (activity of B)

= ${\lambda }_A{N}_A-{\lambda }_B{N}_B,$ ....(ii)

Since each decay of A increases the number of atoms of B

$\frac{d{N}_c}{dt}={\lambda }_B{N}_B$ ....(iii)

Solving equation (i), we get

N_A (t) = N₀ $e^{-{\lambda }_{A}t}$ ....(iv)

Substituting this in equation (ii), we get. the solution,

N_B (t) = $\frac{{\lambda }_A{N}_0{e}^{-{\lambda }_{A}t}}{{\lambda }_A+{\lambda }_B}+C_1{e}^{-{\lambda }_{A}t}$ ......(v)

Putting the initial value (t = 0) of N_B, we get C₁^, and substituting this value of C₁ we get the expression for N_B :

N_B (t) = $\frac{{\lambda }_A{N}_0}{-{\lambda }_A+{\lambda }_B}\left[\frac{1-e^{-{\lambda }_{A}t}}{{\lambda }_A}-\frac{1-e^{-{\lambda }_{B}t}}{{\lambda }_B}\right]$ .......(vi)

Substitution the value of N_B in equation (iii) and solving, we get

N_C = $\frac{{\lambda }_A{\lambda }_B{N}_0}{-{\lambda }_A+{\lambda }_B}\left[\frac{1-e^{-{\lambda }_{A}t}}{{\lambda }_A}-\frac{1-e^{-{\lambda }_{B}t}}{{\lambda }_B}\right]$ ......(vii)

Using these values of N_A, N_B and N_C. the activity can be calculated, if required.