Activity of a radioactive sample

Answer

Activity

Activity (A) of a radioactive material/sample is the number of disintegrations per second for the sample. we can also define activity as the the number of unstable atomic nuclei that decay per second in a given sample. Equivalently, it can be looked at as the magnitude of decay rate. Activity also represents the number of radiations emitted i.e the number of alpha,beta and gamma rays emitted that are in proprtion to the number of disintegrations.

The activity of a radioactive nuclide depend on two parameters; The quantity of radioactive material and its half-life. Indeed, the more the atomic nuclei of the radioisotope the higher the activity in terms if disintegrations per unit time, on the other hand, for a given number of radioactive nuclei, half-life is inversely proportional to its activity (equation 1); nuclides with longer half-lives have lower activities. As a radioactive nuclide decays to daughter nuclide, its quantity and activity gradually decrease with time.

Units of activity

Activity has the units disintegrations per second (dps). Activity can also be reported in other units such as Curie (Ci) and Becquerel (Bq). These two other units can be expressed in terms of disintegrations per second (dps) using the interconversions illustrated below.

       $$\begin{gathered} 1 Curie\mathit{ }\mathit{\left(}Ci\mathit{\right)}\mathit{ }\mathit{=}\mathit{ }\mathit{3}\mathit{.}\mathit{7}\mathit{ }\mathit{\times }\mathit{ }{\mathit{10}}^{\mathit{10}}\mathit{ }dps \\ 1 Curie \left(Ci\right) = \mathit{3}\mathit{.}\mathit{7}\mathit{ }\mathit{\times }\mathit{ }{\mathit{10}}^{\mathit{10}}\mathit{ }Bq \\ Becquerel\mathit{ }\mathit{\left(}Bq\mathit{\right)}\mathit{=}\mathit{1}\mathit{ }dps \\ \mathit{1}\mathit{ }Ci\mathit{ }\mathit{=}\mathit{ }\mathit{3}\mathit{.}\mathit{7}\mathit{ }\mathit{\times }\mathit{ }{\mathit{10}}^{\mathit{10}}\mathit{ }dps \end{gathered}$$

Since activity is the number of disintegrations per unit time,  we can express activity in terms of the total number of radioactive nuclei N and the decay constant λ,

$A=\lambda N$ or $A$= $\frac{0.693}{t_{1/2}}N$                                                                                             1 

but we also know that the equation for the exponential decay is written as;

$N=N_o{e}^{-\lambda t}$                                                                                                               2       

Using equations 1 and 2, we obtaing an equation of actvity as $A=\lambda N=N_o{e}^{-\lambda t}$ , where $N_{o }$ is the number of nuclei at $t$=0.    

Similarly, we can write exponential decay equation 2 interms of activity;

$A$ = $A_o$$e^{-\lambda t}$                                                                                                             3

Equation 3 gives the number of nuclei that actually decayed is identical to the equation;

 $A$ = $A_o$${\left(\frac{1}{2}\right)}^n$                                                                                                          4

Where n is the number of half-lives. In these  two equations (3 and 4), $e^{-\lambda t}$ and ${\left(\frac{1}{2}\right)}^n$ are identical and both represents the fraction of the radioactive nuclei that remained, it is also called the decay factor.

Simplifying quation 4 yields the equation; 

$A$ = $\frac{A_o}{2^n}$ or $\frac{A}{A_o}$=$\frac{1}{2^n}$                                                                                         5

From equation 5, its clear that in n half-lives, the fraction of activity that decayed is given by;

$1-\frac{A}{A_o}$                                                                                                              6

Example

By considering an initial ${10}^9$ atoms of Fluorine-18 (half-life= 110 minutes),

a) How many atoms of this radioisotope will remain undecayed after 220 minutes?

Solution

By shall utilize equation 5,

$\frac{A}{A_o}$= $\frac{1}{2^n}$

$n$ = $\frac{220}{110}$

   = 2 

$A$=$\frac{1}{2^2}$$A_o$

   = $\frac{1}{4}$$\times$${10}^9$

   = $2.5 \times {10}^8 atoms$

b) What is the fraction that remained?

       $\frac{A}{A_o}$= $\frac{1}{2^2}$

             = $\frac{1}{4}$

An alternative method is calculating $e^{-\lambda t}$, $\lambda$ = $\frac{0.693}{110}$, we have $t$ = 110 minutes

$e^{-\lambda t}$ = $e^{-\left(\frac{0.693}{110}\right)\times 220}$

       =$0.25$

       = $\frac{1}{4}$

c) Fraction that decayed

This is given by $1-\frac{A}{A_o}$

                    = $1-\frac{1}{4}$

                    = $\frac{3}{4}$

SPECIFIC ACTIVITY 

Specific activity is defined as the activity per unit mass of atoms of a given radioactive nuclide. it is normally reported in the units Bq/g or Ci/g.

For a given radioactive nuclide, the number of atoms in 1 g, denoted ($N$), is given by the relation

$N$ = $\frac{6.02 \times {10}^{23} atoms/mole}{atomic weight in g/mole}$ $\times$ $1 g$                                                   7

The number of atoms in 1 kg (1000g) of a radioactive nuclide is therefore given by

$N$ = $\left(\frac{6.02 \times {10}^{23} atoms per mole}{A_w}\right) \times 1000 g$, where $A_w$ is atomic weight in g/mole

From equation 7, We can draw a deduction that the mass of radioactive nuclide can be computed using the equation;

$Mass \left(g\right)$ = $\frac{N \times A_w}{N_A}$, where $N$ is the number of atoms per gram, $A_w$ is molar mass in g/mol and                                     $N_A$ is the avogadros constant ($6.02 \times {10}^{23}$)

Specific activity interms of decay constant (λ)and half-life ($t_{1/2}$) 

Specific activity in (Bq/g) can expressed in terms of decay constant using the relation;

$Specific activity (Bq/g)$ = $\frac{\lambda N}{\left({\displaystyle \frac{A_{w}N}{N_A}}\right)}$

                                =    $\frac{\lambda N}{A_{w}N} \times N_A$

                                = $\frac{\lambda N_A}{A_w}$                                                                 8

We also know that $\lambda$ = $\frac{\ln 2}{t_{1/2}}$, subsituting this in equation 8, we obtain an equation of specific activity with half-life component;

$Specific activity (Bq/g)$ = $\frac{N_{A }\ln 2}{A_w t_{1/2}}$                                                    9

equation 9 can be simplified by using the numerical values of $N_A$ and $\ln 2$, equation 9 thjerefore reduces to;

 $Specific activity (Bq/g)$ = $\frac{4.17 \times { 10}^{23}}{A_w t_{1/2}}$,                                            10

The units of half-life in equation 10 should strictly be in seconds. But if the units of half-life is in years then equation 10 can further be sinplified to yield;

 $Specific activity (Bq/g)$ = $\frac{1.32 \times {10}^{16}}{A_w t_{1/2}}$                                             11

Examples

1. Given that the half-life of Cs-137 is 30.17 years, calculate its specific activity.

Solution

We solve this problem by using equation 11, because the units of half-life is in years, remember that mass number is $1\%$ of molar mass, we therefore use it as molar mass.

$Bq/g$ = $\frac{1.32 \times {10}^{16}}{137 \times 30.17}$

          = $3.19 \times {10}^{12}$

2. Calculte the half-life of a radioactive nuclide (Mass number 87) that has a specific activity of $3.2 \times {10}^6 Bq/kg$.

Solution

Equation 9 can be rearrange to give half-life in terms of other quantities;

$t_{1/2}$ = $\frac{N_A \ln 2}{Bq/g \times A_w}$,                                                                          12

Equation 12 gives half-life in seconds and if need be, we convert to years

We first convert activity to to the units Bq/g by multiplying by 1000 to yield 3200 Bq/g

$t_{1/2}$ = $\frac{6.023 \times {10}^{23} \times 0.693}{3200 \times 87}$

       = $1.499 \times {10}^{18} seconds$

       = 47 billion years

REFERENCES

L'Annunziata, M. F. (Ed.). (2012). Handbook of radioactivity analysis. Academic press

Wahl, A. C., & Bonner, N. A. (1951). Radioactivity applied to chemistry (Vol. 72, No. 5, p. 405). LWW.

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