6.2 Simultaneous decay modes of radioactive elements

If an element disintegrates with two or more decay modes simultaneously into different daughter nuclei with different decay constants ${\lambda _1},\;{\lambda _2},\;{\lambda _3}{\text{ etc}}{\text{.,}}$ for each mode, then the effective decay constant of the parent nuclei can be given as

$${\lambda _{eff}} = {\lambda _1} + {\lambda _2} + {\lambda _3} + ...$$

Co sider a radioactive series where each element decays into its daughter nuclei until a stable element appears.

$${A_1}\left( {{\lambda _1}} \right) \to {A_2}\left( {{\lambda _2}} \right) \to {A_3}\left( {{\lambda _3}} \right) \to ...$$

Production rate of ${A_2} = {\lambda _1}{N_1}$

Decay rate of ${A_2} = {\lambda _2}{N_2}$

So, we can write rate equation as,

$$\begin{equation} \begin{aligned} \,\frac{{d{N_2}}}{{dt}} = {\lambda _1}{N_1} - {\lambda _2}{N_2} \\ \frac{{d{N_2}}}{{dt}} + {\lambda _2}{N_2} = {\lambda _1}{N_0}{e^{ - \lambda t}} \\\end{aligned} \end{equation} $$

Solving this linear differential equation we get

$${\text{ }}{N_2} = \frac{{{\lambda _1}{N_0}}}{{\left( {{\lambda _1} - {\lambda _2}} \right)}}\left( {{e^{ - {\lambda _2}t}} - {e^{ - {\lambda _1}t}}} \right)$$

Due to disintegration of $A_1$, $A_2$ is being formed.

As the amount of $A_1$ is decreased and that of $A_2$ is increased, decay rate of $A_2$ increases and that of $A_1$ decreases.

After a time $t$ when both the decay rates become equal, the element $A_2$ will be said to be in radioactive equilibrium when the yield of the radioactive nuclide $A_2$ is maximum.

Question 15. A factory produces a radioactive substance $A$ at a constant rate $R$ which decays with a decay constant $\lambda $ to form a stable substance.

Find

(a) the number of nuclei of $A$

(b) number of nuclei of $B$, at any time $t$ assuming the production of $A$ starts at $t=0$.

(a) \[Factory\xrightarrow[{const.\;rate}]{R}A\xrightarrow[{decay}]{\lambda }B\]

Let $N$ be the number of nuclei of $A$ at any time $t$,

$$\frac{{dN}}{{dt}} = R - \lambda {N_A}$$

Integrating with proper limits we get, $$\begin{equation} \begin{aligned} \int\limits_0^{{N_A}} {\frac{{dN}}{{R - \lambda {N_A}}}} = \int\limits_0^t {dt} \\ \left[ {\ln \left( {R - \lambda {N_A}} \right)} \right]_0^{{N_A}} = \left[ t \right]_0^t \\ \ln \left( {\frac{{R - \lambda {N_A}}}{R}} \right) = - \lambda t \\ \left( {\frac{{R - \lambda {N_A}}}{R}} \right) = {e^{ - \lambda t}} \\ {N_A} = \frac{R}{\lambda }\left( {1 - {e^{ - \lambda t}}} \right) \\\end{aligned} \end{equation} $$

(b) Number of nuclei $B$ at any time $t$,

$$\begin{equation} \begin{aligned} {N_B} = Rt - {N_A}\quad \; \\ {N_B}\; = Rt - \frac{R}{\lambda }\left( {1 - {e^{ - \lambda t}}} \right) \\ {N_B} = R\left( {\lambda t - 1 + {e^{ - \lambda t}}} \right) \\\end{aligned} \end{equation} $$

Question 16. Nuclei of radioactive element $A$ are being produced at a constant rate $\alpha $. The element has a decay constant $\lambda $. At time $t=0$, there are $N_0$ nuclei of the element.

(a) Calculate the number of nuclei of $A$ at time $t$.

(b) If $\alpha = 2{N_0}\lambda $, calculate the number of nuclei of $A$ after one half-life time of $A$ and also the limiting value of $N$ as $t \to \infty $.

(a) The rate of formation of radioactive nuclei is $\alpha $.

Rate of decay of radioactive nuclei is $\lambda N$.

Therefore, $$\begin{equation} \begin{aligned} \frac{{dN}}{{dt}} = \alpha - \lambda N \\ \frac{{dN}}{{\alpha - \lambda N}} = dt\quad ...(i) \\\end{aligned} \end{equation} $$

At $t = 0$, the number of nuclei is $N=N_0$.

So, integrating with proper limits we get,

$$\begin{equation} \begin{aligned} \int\limits_{{N_0}}^N {\frac{{dN}}{{\alpha - \lambda N}}} = \int\limits_0^t {dt} \\ \ln \left( {\frac{{\alpha - \lambda N}}{{\alpha - \lambda {N_0}}}} \right) = - \lambda t \\ \frac{{\alpha - \lambda N}}{{\alpha - \lambda {N_0}}} = {e^{ - \lambda t}} \\ N = \frac{\alpha }{\lambda }\left( {1 - {e^{ - \lambda t}}} \right) + {N_0}{e^{ - \lambda t}}...(ii) \\\end{aligned} \end{equation} $$

(b) Given, $\alpha = 2{N_0}\lambda $

$${T_{\frac{1}{2}}} = \frac{{\ln 2}}{\lambda } = \frac{{0.693}}{\lambda }$$

$$\begin{equation} \begin{aligned} N = \frac{{2{N_0}\lambda }}{\lambda }\left( {1 - {e^{ - 0.693}}} \right) + {N_0}{e^{ - 0.693}}\quad \quad \\ N = {N_0}\left( {2 - {e^{ - 0.693}}} \right) \\ N = {N_0}\left( {2 - 0.5} \right) \\ N = 1.5{N_0} \\\end{aligned} \end{equation} $$

When $t \to \infty $, from equation $(ii)$ we can write,

$$N = \frac{\alpha }{\lambda }$$