Calculation of Radioactive Equilibrium

TL;DR

Radioactivity at any time t

\[\begin{align*} \alpha (t) &= \lambda n(t) \\ &= \alpha_0 e^{-\lambda t} \\ &= \alpha_0 e^{-0.693t/T_{1/2}} \end{align*}\]

Relationship between decay constant, half-life, and mean lifetime

\[\begin{align*} T_{1/2}&=\frac {\ln 2}{\lambda} = \frac {0.693}{\lambda} \\ \\ \overline{t}&=\frac {1}{\lambda} \\ &=\frac {T_{1/2}}{0.693}=1.44T_{1/2} \end{align*}\]

Decay Constant

• The probability that a nucleus will decay per unit time
• A constant that is independent of time and determined only by the nuclide
• Denoted by the symbol $\lambda$

Radioactivity

If the number of nuclei that have not yet decayed at time $t$ is n(t), then on average $\lambda n(t)$ nuclei decay during the interval $dt$ between times $t$ and $t+dt$. This decay rate is called the radioactivity of the sample and is denoted by the symbol $\alpha$. Therefore, the radioactivity at any time $t$ is:

\[\alpha (t)=\lambda n(t) \tag{1}\]

Units of Radioactivity

Curie (Ci)

• Traditionally used unit before the becquerel
• Radioactivity of 1g of radium-226
• $3.7\times 10^{10}$ nuclear decays per second ($3.7\times 10^{10}\text{Bq}$)

Becquerel (Bq)

• International Standard (SI) unit
• One nuclear decay per second
• $1 \text{Bq} = 2.703\times 10^{-11}\text{Ci} = 27\text{pCi}$

Calculation of Radioactivity Change Over Time

Since $\lambda n(t)$ nuclei decay during time $dt$, the decrease in the number of nuclei remaining in the sample during $dt$ can be expressed as:

\[-dn(t)=\lambda n(t)dt\]

Integrating this gives:

\[n(t)=n_0e^{-\lambda t} \tag{2}\]

Multiplying both sides by $\lambda$, the radioactivity becomes:

\[\alpha (t)=\alpha_0e^{-\lambda t} \tag{3}\]

Since radioactivity halves during the half-life:

\[\alpha (T_{1/2})=\alpha_0/2\]

Substituting this into equation (3):

\[\alpha_0/2=\alpha_0e^{-\lambda T_{1/2}}\]

Taking the logarithm of both sides and solving for the half-life $T_{1/2}$:

\[T_{1/2}=\frac {\ln 2}{\lambda}=\frac {0.693}{\lambda} \tag{4}\]

Solving this for $\lambda$ and substituting into equation (3):

\[\alpha (t)=\alpha_0e^{-0.693t/T_{1/2}} \tag{5}\]

Equation (5) is often more convenient for radioactive decay calculations than equation (3), as half-life values are more commonly given than decay constants.

The mean lifetime $\overline{t}$ of a radioactive nucleus is the reciprocal of the decay constant:

\[\overline{t}=1/\lambda\]

From equation (3), we can see that during one mean lifetime, the radioactivity falls to $1/e$ of its initial value. From equation (4), the following relationship holds between mean lifetime and half-life:

\[\overline{t}=\frac {T_{1/2}}{0.693}=1.44T_{1/2} \tag{6}\]

※ Derivation of Mean Lifetime $\overline{t}$

\[\begin{align*} \overline{t}&=\frac {\int_0^\infty t\alpha(t)}{\int_0^\infty t} = \frac {\int_0^\infty t\alpha(t)}{n_0} \\ &= \frac {\int_0^\infty n_0 \lambda te^{-\lambda t}}{n_0} \\ &= \int_0^\infty \lambda te^{-\lambda t} \\ &= \left[-te^{-\lambda t}\right]_0^\infty +\int_0^\infty e^{-\lambda t} \\ &=\left[-\frac {1}{\lambda} e^{-\lambda t}\right]_0^\infty \\ &=\frac {1}{\lambda} \end{align*}\]

Example: Radioactive Decay Chain 1

Assume that a radioactive nuclide is produced at a rate of $R$ atoms/s. These nuclei undergo radioactive decay as soon as they are formed. Calculate the radioactivity of this nuclide at any time t.

flowchart LR
	Start[?] -- R --> A[Mathematical Model]
	A -- α --> End[?]

1. Setting up the Model

\[\text{Rate of change of nuclide} = \text{Production rate} - \text{Loss rate}\]

Expressed in mathematical notation:

\[dn/dt = -\lambda n + R\]

2. General Solution

Move all terms related to $n$ to the left side and multiply both sides by $e^{\lambda t}$:

\[\frac {dn}{dt} + \lambda n = R\] \[e^{\lambda t}\frac {dn}{dt} + \lambda e^{\lambda t}n = Re^{\lambda t}\]

Since $\lambda e^{\lambda t}=\frac {d}{dt} e^{\lambda t}$, we can rearrange as follows:

\[e^{\lambda t}\frac {dn}{dt}+\left(\frac {d}{dt} e^{\lambda t}\right)n = Re^{\lambda t}\]

Integrating both sides gives the general solution:

\[e^{\lambda t}n=\frac {R}{\lambda}e^{\lambda t}+c\] \[n=ce^{-\lambda t}+\frac {R}{\lambda}\]

3. Particular Solution

Let’s say the number of this nuclide at $t=0$ is $n_0$ and find the value of the constant $c$:

\[n(0)=c+\frac {R}{\lambda}=n_0\] \[c=n_0-\frac {R}{\lambda}\]

Therefore, the particular solution for the given situation is:

\[n = n_0e^{-\lambda t}+\frac {R}{\lambda}(1-e^{-\lambda t}) \tag{7}\]

We can find the radioactivity of this nuclide by multiplying both sides of the above equation by $\lambda$:

\[\alpha = \alpha_0e^{-\lambda t}+R(1-e^{-\lambda t}) \tag{8}\]

That is, as $t\to\infty$, it converges to $\alpha_{\text{max}}=R$, $n_{\text{max}}=R/\lambda$.

Example: Radioactive Decay Chain 2

Calculate the radioactivity of radioactive nuclide B in the following decay chain:

flowchart LR
	A --> B
	B --> C

1. Setting up the Model

\[\text{Rate of change of B nuclei} = \text{Production rate from A decay} - \text{Decay rate of B to C}\] \[\frac {dn_B}{dt} = -\lambda_B n_B + \lambda_A n_A\]

Substituting equation (2) for $n_A$, we get the following differential equation for $n_B$:

\[\frac {dn_B}{dt} = -\lambda_B n_B + \lambda_A n_{A0}e^{-\lambda_A t} \tag{9}\]

2. General Solution

To solve the differential equation, move all terms related to $n_B$ to the left side and multiply both sides by $e^{\lambda_B t}$:

\[\frac {dn_B}{dt} + \lambda_B n_B = n_{A0}\lambda_A e^{-\lambda_A t}\] \[e^{\lambda_B t}\frac {dn_B}{dt} + \lambda_B e^{\lambda_B t}n_B = n_{A0}\lambda_A e^{(\lambda_B-\lambda_A)t}\]

Since $\lambda_B e^{\lambda_B t}=\frac {d}{dt} e^{\lambda_b t}$, we can rearrange as follows:

\[e^{\lambda_B t}\frac {dn_B}{dt} + \left(\frac {d}{dt} e^{\lambda_B t}\right)n_B = n_{A0}\lambda_A e^{(\lambda_B-\lambda_A)t}\]

Integrating both sides:

\[e^{\lambda_B t}n_B = \frac {n_{A0}\lambda_A}{\lambda_B-\lambda_A}e^{(\lambda_B-\lambda_A)t}+c\]

Dividing both sides by $e^{\lambda_B t}$ gives the general solution:

\[n_B = \frac {n_{A0}\lambda_A}{\lambda_B-\lambda_A}e^{-\lambda_A t}+ce^{-\lambda_B t}\]

3. Particular Solution

Let’s say the number of element B at $t=0$ is $n_{B0}$ and find the value of the constant $c$:

\[n_B(0)=\frac {n_{A0}\lambda_A}{\lambda_B-\lambda_A}+c=n_{B0}\] \[c=n_{B0}-\frac{n_{A0}\lambda_A}{\lambda_B-\lambda_A}\]

Therefore, the particular solution for the given situation is:

\[n_B = n_{B0}e^{-\lambda_B t} + \frac {n_{A0}\lambda_A}{\lambda_B - \lambda_A} (e^{-\lambda_A t} - e^{-\lambda_B t}) \tag{10}\] \[\therefore \alpha_B = \alpha_{B0} e^{-\lambda_B t} + \frac {\alpha_{A0}\lambda_A}{\lambda_B - \lambda_A} (e^{-\lambda_A t} - e^{-\lambda_B t}) \tag{11}\]