Simultaneous And Series Disintegration

Simultaneous Decay

Simultaneous decay, also known as simultaneous disintegration, occurs when a single unstable nucleus decays into two or more different products at the same time. This process is less common compared to series decay, where a nucleus undergoes a series of decays, but it provides valuable insights into the behaviour of certain isotopes and the interactions between subatomic particles.

As we know due to radioactive disintegration, a radio nuclide transforms into its daughter nucleus. Depending on the nuclear structure and its instability, a parent nucleus may undergo either $\alpha-$ or $\beta$-emission. Sometimes a parent nucleus may undergo both types of emission simultaneously.

If an element decays to different daughter nuclei with different decay constants etc. for each decay mode, then the effective decay constant of the parent nuclei can be given as

$\lambda_{e f f}=\lambda_1+\lambda_2+\lambda_3, \ldots$

Similarly, a radioactive element with decay constant $\lambda$ which decays by both $\alpha-$ and $\beta-$ decays given that the probability for an $\alpha$--emission is $\mathrm{P}_1$ and that for $\beta-$ emission is $\mathrm{P}_2$ the decay constant of the element can be split for individual decay modes. Like in this case the decay constants for $\alpha$ - and $\beta-$ decay separately can be given as

$\begin{aligned} & \lambda_\alpha=P_1 \lambda \\ & \lambda_\beta=P_2 \lambda\end{aligned}$

Series Decay

Series decay, also known as decay chains or radioactive series, refers to a sequence of radioactive decays where a parent isotope decays into a daughter isotope, which in turn decays into another isotope, and this process continues until a stable isotope is formed. This type of decay is fundamental in understanding the transformation and stability of elements over time.

Accumulation of Radioactive Elements in Radioactive Series

The accumulation of radioactive elements in a radioactive series, also known as a decay chain, occurs as a parent isotope decays into a series of daughter isotopes until a stable isotope is reached. Each step in the decay chain involves a specific type of radioactive decay (alpha, beta, or gamma decay) and results in the formation of a new element.

A radioactive element decays into its daughter nuclei until a stable element appears. Consider a radioactive series.

$A_1 \xrightarrow{\lambda_1} A_2 \xrightarrow{\lambda_2} A_3 \xrightarrow{\lambda_3} \ldots$

A radioactive element $A_1$ disintegrates to form another radioactive element $A_2$ which in turn disintegrates to another element $A_3$ and so on. Such decays are called Series or Successive Disintegration.

Here, the rate of disintegration of $A_1=$ Rate of formation of $A_2$
$
\begin{gathered}
\frac{-d N_{A 1}}{d t}=\frac{d N_{A 2}}{d t}=\lambda_1 N_{A 1} \\
\frac{-d N_{A 2}}{d t}=\frac{d N_{A 3}}{d t}=\lambda_2 N_{A 2} \\
\frac{d N_{A 1}}{d t}=-\lambda_1 N_{A 1} \\
\frac{d N_{A 2}}{d t}=-\lambda_2 N_{A 2}
\end{gathered}
$

Therefore, net formation of $A_2=$ Rate of disintegration of $A_1$ - Rate of disintegration of $A_2$
$
=\lambda_1 N_{A 1}-\lambda_2 N_{A 2}
$

If the rate of disintegration of $A_1$ becomes equal to the Rate of disintegration of $A_{2 \text {, }}$ then it is called Radioactive equilibrium. So the equation becomes -
$
\Rightarrow \frac{\lambda_1}{\lambda_2}=\frac{N_{A 2}}{N_{A 1}}=\frac{T_{a v g 2}}{T_{a v g 1}}=\frac{\left(T_{\frac{1}{2}}\right)_2}{\left(T_{\frac{1}{2}}\right)_1}
$

Recommended Topic Video

Solved Examples Based on Simultaneous and Series Disintegration

Example 1: The half-life of radioactive substances is 1620 years and 405 years for $\alpha$ emission and $\beta$ emission respectively. Time (in years) during which half of the sample will decay if both $\alpha$ and $\beta$ emission are simultaneous is:

1) 324

2) 405

3) 1620

4) 2025

Solution:

The effective decay constant is when nuclei decay into more than one product at one time.

$
\begin{aligned}
\lambda & =\lambda_1+\lambda_2=\frac{0.693}{T}=\frac{0.693}{T_1}+\frac{0.693}{T_2} \\
\frac{1}{T} & =\frac{1}{T_1}+\frac{1}{T_2} \\
\frac{1}{T} & =\frac{1}{1620}+\frac{1}{405} \\
\frac{1}{T} & =\frac{5}{1620}
\end{aligned}
$

OR $T=324$ years
Hence, the answer is option(1).

Example 2: A radioactive nuclei with a decay constant of $0.5 / \mathrm{s}$ is being produced at a constant rate of 100 nuclei/s. If at $t=0$, there were no nuclei, the time when there are 50 nuclei is :
1) 1 s
2) $2 \ln \left(\frac{4}{3}\right) \mathrm{s}$
3) $\ln 2 s$
4) $\ln \left(\frac{4}{3}\right) s$

Solution:

At any time
$
t: \frac{d N}{d t}=N_o-\lambda N
$
$N_o=$ Rate of production
$
\begin{aligned}
& -\lambda N=\text { Rate of decay } \\
& { }_o^N \int \frac{d N}{N_o-\lambda N}=\int_o^t d t \\
& \text { or } \frac{-1}{\lambda} \ln N_o-\lambda N \int_o^N=t \\
& \text { or } \ln \left(\frac{N_o-\lambda N}{N_o}\right)=-\lambda t \\
& 1-\frac{\lambda N}{N_o}=e^{-\lambda t} \\
& \text { or } N=\frac{N_o}{\lambda}\left(1-e^{-\lambda t}\right) \\
& \text { or }
\end{aligned}
$
given $N=50, N_o=100, \lambda=0.5$
$
\begin{aligned}
& \Rightarrow 50=\frac{100}{0.5}\left(1-e^{-\lambda t}\right) \\
& \frac{1}{4}=1-e^{-0.5 t} \text { or } \frac{3}{4}=e^{-0.5 t}
\end{aligned}
$
taking log;
$
\ln \left(\frac{4}{3}\right)=0.5 t
$
or $t=2 \ln \frac{4}{3} s$

Hence, the answer is the option (2).

Example 3: A radioactive nucleus decays by two different processes. The half-life of the first process is 5 minutes and that of the second process is 30 s . The effective half-life of the nucleus is calculated to be $\frac{\alpha}{11} \mathrm{~s}$. The value of $\alpha$ is $\qquad$ .

1) 300

2) 400

3) 500

4) 600

Solution:

$\begin{aligned} & \frac{\mathrm{dN}}{\mathrm{dt}}=-\left(\lambda_1+\lambda_2\right) \mathrm{N} \\ & \lambda_{\text {eq }}=\lambda_1+\lambda_2 \\ & \frac{1}{\mathrm{t}_{\frac{1}{2}}}=\frac{1}{300}+\frac{1}{30}=\frac{11}{300} \\ & \mathrm{t}_{1 / 2}=\left(\frac{300}{11}\right) \mathrm{sec}\end{aligned}$

Hence, the answer is option (1).

Example 4: A radioactive nucleus undergoes a series of decay according to the scheme
$
\mathrm{A} \xrightarrow{\alpha} \mathrm{A}_1 \xrightarrow{\beta} \mathrm{A}_2 \xrightarrow{\alpha} \mathrm{A}_3 \xrightarrow{\gamma} \mathrm{A}_4
$

If the mass number and atomic number of $A$ are 180 and 72 respectively, these numbers of $\mathrm{A}_4$ are:

1) 172, 69

2) 177, 69

3) 171, 69

4) 172, 68

Solution:

Decrease in mass number due to two $\alpha=2 \times 4=8$
Decrease in charge number due to two $\alpha=4$
Increase in charge number due to one $\beta=1$
Net decrease in charge number $=4-1=3$
$\gamma$ affects neither.
$\therefore \quad$ For $A_4, A=180-8=172$
$
Z=72-3=69
$

Hence, the answer is the option (1).

Example 5: The mean lives of a radioactive sample are 30 years and 60 years for $\alpha$ emission and $\beta$-emission respectively. If the sample decays both by $\alpha$-emission and -emission simultaneously, the time after which, only one-fourth of the sample remains:

1) 14 years

2) 20 years

3) 28 years

4) 45 years

Solution:

Here, $\lambda_{(\alpha+\beta)}=\lambda_\alpha+\lambda_\beta$
$
\begin{aligned}
& \frac{1}{\tau}=\frac{1}{\tau_\alpha}+\frac{1}{\tau_\beta} \quad\left(\text { As } \lambda=\frac{1}{\tau}\right) \\
& \Rightarrow \frac{1}{\tau}=\frac{1}{30}+\frac{1}{60}=\frac{1}{20} \\
& \therefore \quad \tau=20 \text { years }
\end{aligned}
$

Now, $\mathrm{T}_{1 / 2}=\ln (2) \tau=13.86$ years
One-fourth of the sample will remain after 2 half-life $=27.72$ years.
Hence, the answer is the option (3).

Summary

Simultaneous and series disintegration are crucial concepts in nuclear physics that describe the decay processes of unstable nuclei. Simultaneous decay involves a single nucleus decaying into multiple products at the same time, providing insights into the behaviour of certain isotopes. In contrast, series decay, or radioactive series, involves a sequence of decays where a parent isotope transforms into a series of daughter isotopes until a stable form is achieved.