Advanced Modern Physics
6.0 Radioactive decay law
6.0 Radioactive decay law
Fundamental laws of Radioactivity:
- Radioactivity is purely a nuclear process. It is not concerned in any way with the electrons orbiting around the nucleus.
- As radioactivity is a nuclear process, it is independent from any chemical property of the element. As we have discussed that radioactive property of an element is only the process concerned with the nucleus of the element, it doesn't affect the electronic configuration of the element. If this element takes part in a chemical reaction, the product formed will also have the radioactive property in the same fraction by which the radioactive atom is present in the molecules of the product.
- As radioactivity is a random process, its study is only possible by laws of probability. In a group of several radioactive elements, which one will disintegrate first is just a matter of chance.
- As radioactivity is a random process, the disintegration density throughout the volume of the radioactive element remains constant. If an element $X$ decays to a daughter nuclide $Y$, then in a given volume of the element, all portions of volume will have the same ratio of number of atoms of $Y$ to that of $X$. Thus, homogeneity is maintained.
Thus, due to randomness the amount of disintegrations per unit volume per second (disintegration density) remains approximately constant in the whole volume of the substance.
The radioactivity of an element is measured in terms of activity.
Activity of a sample of any radioactive element is the rate at which the number of its constituents atoms disintegrate.
If $N$ is the number of nuclei present in a radioactive sample at any instant, then the activity of the sample is given as,
$$\begin{equation} \begin{aligned} {A_c} = - \frac{{dN}}{{dt}} \\ \\\end{aligned} \end{equation} $$
Here, -ve sign suggests that $\left( {\frac{{dN}}{{dt}}} \right)$ is decreasing.
So, $A_c$ is always taken positive. It is measured in $dps$ or disintegrations per second.
SI unit of activity is named after Becquerel, defined as 1 bequeral=1 $Bq$=1 $dps$.
Sometimes traditional units like Curie and Rutherford are also used,
$$\begin{equation} \begin{aligned} 1{\text{ }}Ci = 3.7 \times {10^{10}}{\text{ }}dps \\ 1{\text{ }}Ru = {10^6}{\text{ }}dps \\\end{aligned} \end{equation} $$
Let us consider that at $t = 0$, there are $N_0$ parent atoms in a substance and after a time $t$, $N$ atoms are left decayed.
This implies that in a duration from $t=0$ to $t=t$, ${N_0} - N$ atoms are decayed to their daughter element.
If in further time from $t=t$ to $t = t + dt$, $dN$ more atoms will decay then at anytime time $t=t$,
$A \to $$BB$
$t=0$${N_0}$
$t=N$$\left( {{N_0} - N} \right)$
We can say that activity of the element is given as,
$${A_c} = - \frac{{dN}}{{dt}}{\text{ }}$$
Now, according to radioactive decay law,
$$ - \frac{{dN}}{{dt}} \propto N$$ or $$\frac{{dN}}{{dt}} = - \lambda N...(i)$$
where,
$\lambda $: is the proportionality constant known as decay constant of a decay process.
The decay constant is different for different radioactive substances, but it doesn't depend on the amount of substance and time.
The element having a high value of $\lambda $ have a high value of activity and vice-versa.
Thus, we can say that the decay constant for a radioactive element gives the relative criteria of its stability as well as the rate of reaction.
If the value of $\lambda $ for an element is more, it is more active or relatively less stable than the one having a smaller value of $\lambda $.
SI unit of $\lambda $ is ${s^{ - 1}}$
Integrating within limits we have,
$$\begin{equation} \begin{aligned} \quad \quad \int\limits_{{N_0}}^N {\frac{{dN}}{N}} = - \int\limits_0^t {\lambda dt} \\ \quad \quad \Rightarrow \;\left[ {\ln N} \right]_{{N_0}}^N = - \lambda t \\ \quad \quad \Rightarrow \;\ln \left( {\frac{N}{{{N_0}}}} \right) = - \lambda t\quad {\text{or}}\quad {N = {N_0}{e^{ - \lambda t}}}...(ii) \\\end{aligned} \end{equation} $$
Multiplying $\lambda $ on each side of $(ii)$ we have,
$${A_c} = {A_{{c_0}}}{e^{ - \lambda t}}...(iii)$$
where ${A_{{c_0}}}$ is the initial activity of the substance at $t=0$.
And number of nuclei decayed, $$\begin{equation} \begin{aligned} N' = {N_0} - N \\ N' = {N_0} - {N_0}{e^{ - \lambda t}} \\\end{aligned} \end{equation} $$ or $$N' = {N_0}\left( {1 - {e^{ - \lambda t}}} \right)$$
Half-life $\left( {{T_{\frac{1}{2}}}} \right)$
It is defined as the time duration in which half of the total number of nuclei will decay or be left decayed.
Let at half life, $t = {T_{\frac{1}{2}}}$ number of decayed nucleus is, $N = \frac{{{N_0}}}{2}$.
As we know, $$N = {N_0}{e^{ - \lambda t}}$$ So, $$\begin{equation} \begin{aligned} \frac{{{N_0}}}{2} = {N_0}{e^{ - \lambda {T_{\frac{1}{2}}}}}\quad \quad \\ \left( {\frac{{{N_0}}}{2} \times \frac{1}{{{N_0}}}} \right) = {e^{ - \lambda {T_{\frac{1}{2}}}}}\quad \quad \\ \frac{1}{2} = {e^{ - \lambda {T_{\frac{1}{2}}}}} \\\end{aligned} \end{equation} $$ Taking logarithm on both sides, $$\begin{equation} \begin{aligned} \ln \left( {\frac{1}{2}} \right) = \ln \left( {{e^{ - \lambda {T_{\frac{1}{2}}}}}} \right) \\ \ln \left( {\frac{1}{2}} \right) = - \lambda {T_{\frac{1}{2}}} \\ \ln \left( 2 \right) = \lambda {T_{\frac{1}{2}}} \\ {T_{\frac{1}{2}}} = \frac{{\ln \left( 2 \right)}}{\lambda } \\\end{aligned} \end{equation} $$
Number of nuclei present after $n$ half-life i.e., after a time, $t = n{T_{1/2}}$
$$\begin{equation} \begin{aligned} N = {N_0}{e^{ - \lambda n{T_{\frac{1}{2}}}}} \\ N = {N_0}{e^{ - \lambda n\frac{{\ln \left( 2 \right)}}{\lambda }}}\quad \left( {As,\;{T_{\frac{1}{2}}} = \frac{{\ln \left( 2 \right)}}{\lambda }} \right) \\ N = {N_0}{e^{\ln \left( {\frac{1}{{{2^n}}}} \right)}} \\ N = \frac{{{N_0}}}{{{2^2}}} \\\end{aligned} \end{equation} $$
Average life $\left( {{T_{avg}}} \right)$
$$\begin{equation} \begin{aligned} {T_{avg}} = \frac{{{\text{Sum of ages of all the nuclei}}}}{{{N_0}}} \\ {T_{avg}} = \frac{{\int\limits_0^\infty {{N_0}{e^{ - \lambda t}}dt} }}{{{N_0}}} \\ {T_{avg}} = \frac{1}{\lambda } \\\end{aligned} \end{equation} $$
It is also known as mean life or time constant.
Question 13. The half-life of a radioactive nuclide is 20 hours. What fraction of original activity will remain after 40h?
Solution:
40 hours means two half-lives. Thus,
$$\begin{equation} \begin{aligned} A = \frac{{{A_0}}}{{{2^2}}} \\ A = \frac{{{A_0}}}{4}\quad \\ \frac{A}{{{A_0}}} = \frac{1}{4} \\\end{aligned} \end{equation} $$
So, one fourth of the original activity will remain after 40 hours.
Specific activity
The activity per unit mass is called specific activity.
Question 14. The half-life of ${}^{198}Au$ is 2.7 days. Calculate
(a) the decay constant
(b) the average-life
(c) the activity of 1.00 mg of ${}^{198}Au$.
(Take atomic weight of $^{198}Au$ to be $198\;g\;mo{l^{ - 1}}$).
Solution:
(a). The half-life and decay constant are related as,
$$\begin{equation} \begin{aligned} {T_{\frac{1}{2}}} = \frac{{\ln 2}}{\lambda } \\ {T_{\frac{1}{2}}} = \frac{{0.693}}{\lambda } \\ \lambda = \frac{{0.693}}{{{T_{\frac{1}{2}}}}} \\ \lambda = \frac{{0.693}}{{2.7{\text{ }}days}} \\ \lambda = \frac{{0.693}}{{2.7 \times 24 \times 3600{\text{ }}s}} \\ \lambda = 2.9 \times {10^{ - 6}}{\text{ }}{s^{ - 1}} \\\end{aligned} \end{equation} $$
(b). The average life is, $${T_{avg}} = \frac{1}{\lambda } = 3.9{\text{ days}}$$
(c). The activity is $A = \lambda N$.
Now, $198\ g$ of ${}^{198}Au$ has $6 \times {10^{23}}$ atoms.
The number of atoms in $1.00\ mg$ of ${}^{198}Au$ is,
$$\begin{equation} \begin{aligned} N = 6 \times {10^{23}} \times \frac{{1.0{\text{ }}mg}}{{198{\text{ }}g}} \\ N = 3.03 \times {10^{18}} \\\end{aligned} \end{equation} $$
So, $$\begin{equation} \begin{aligned} A = \lambda N \\ A = \left( {2.9 \times {{10}^{ - 6}}{\text{ }}{s^{ - 1}}} \right)\left( {3.03 \times {{10}^{18}}} \right) \\ A = 8.8 \times {10^{12}}{\text{ disintegrations}} \\\end{aligned} \end{equation} $$ As, ${\text{1}}\;Ci = 3.7 \times {10^{10}}\;{\text{disintegrations per second}}$ Therefore, $$\begin{equation} \begin{aligned} A = \frac{{8.8 \times {{10}^{12}}}}{{3.7 \times {{10}^{10}}}}{\text{ }}Ci \\ A = 240{\text{ }}Ci \\\end{aligned} \end{equation} $$
