Chapter 4, Problem 4.3P

To determine

To calculate:

The counting rates of ${}^{131}I\text{ and}{}^{198}Au$ over 16-day period at time $t=0,t=3,t=8,t=\text{16}$ . Also, write the equation of curve for total count rate versus time by plotting graph of daily observation on semi-10 g paper.

Answer to Problem 4.3P

Counting rates of ${}^{131}I\text{ and}{}^{198}Au$ are as below:

  $\begin{array}{llll}\text{Day}\hfill & \text{Total count rate}\hfill & \%{}^{\text{198}}Au\hfill & \%{}^{\text{131}}I\hfill \\ 0\hfill & 56.6\hfill & 65\hfill & 35\hfill \\ 3\hfill & 32.4\hfill & 52\hfill & 48\hfill \\ 8\hfill & 14.6\hfill & 32\hfill & 68\hfill \\ 18\hfill & 5.6\hfill & 11\hfill & 89\hfill \end{array}$

The equation of curve for count rate versus time is:

  $A\left(t\right)=36.8e^{-0.257t}+219.8e^{-0.086t}$

Explanation of Solution

Given Information:

Volume of solution $=1ml$

Activity of ${}^{198}Au=370Bq$

Activity of ${}^{131}I=185Bq$

Duration $=16days$

Formula used:

  $A={\left(A_0{e}^{-}{}^{\lambda t}\right)}_{198Au}+{\left(A_0{e}^{-\lambda t}\right)}_{131I}$

Graph of count rate versus time is as below:

We know that ${}^{198}Au$ emits $99.5\%\gamma$ of total transformation having half-life = $2.7$ days.

While ${}^{131}I$ emits at $107\%\gamma$ of total transformation having half-life = $8.05$ days.

  $\begin{array}{l}Att=0,{\text{For}}^{198}Au,\hfill \\ 370\times 99.5/100\times 10/100=36.8\text{ count/sec}\hfill \\ \text{For}{}^{\text{131}}\text{I},\hfill \\ 185\times 107/100\times 10/100=19.8\text{ count/sec}\hfill \end{array}$

Thus, total number of counts= $36.8+19.8=56.6$

Hence, ratio of counts at t=0

  $\begin{array}{l}{\text{For}}^{198}Au=36.8/56.6=0.65=65\%\hfill \\ {\text{For}}^{131}I=19.8/56.6=0.35=35\%\hfill \end{array}$

  $\begin{array}{l}\text{At }t\text{=3,}\text{For}{}^{198}Au,\hfill \\ \lambda =0.693/2.7=0.257d^{-1}\hfill \\ A=A_0{e}^{-\lambda t}=36.8\times e^{-0.257\times 3}=17\text{count/sec}\hfill \end{array}$

Where,

  $\begin{array}{l}{A}_0=36.8\\ \lambda =0.257\\ t=3\end{array}$

For 131I

  $\begin{array}{l}\lambda =0.693/8.05=8.6\times {10}^{-}{}^2{d}^{-1}\\ A=A_0{e}^{-\lambda }{}^t=19.8\times e^{-8.6\times }{}^{{10}^{-2}\times 3}\\ =15.3\text{count/sec}\end{array}$

Where,

  $\begin{array}{l}{A}_0=19.8\\ \lambda =8.6\times {10}^{-2}\\ t=3\end{array}$

  $\begin{array}{l}Thus,totalnumberofcount\\ \begin{array}{l}Henceratioofcountsatt=3\hfill \\ Fo{r}^{198}Au=17/32.4=0.52=52\%\hfill \\ Fo{r}^{131}I=15.3/32.4=0.48=48\%\hfill \end{array}\end{array}$

  $Att=8$

  $\begin{array}{l}For{}^{198}Au\hfill \\ \lambda =0.257d^{-1}\hfill \\ A_0=36.8\hfill \\ t=8\hfill \end{array}$

  $A=A_0{e}^{-\lambda t}=36.8\times e^{-0.257\times 8}=4.7\text{ count/sec}$

  $\begin{array}{l}\text{For}{}^{131}I,\\ \lambda =8.6\times {10}^{-2}\\ A_0=19.8\\ t=8\end{array}$

  $A=A_0{e}^{-\lambda t}=19.8\times e^{-8.6}\times {10}^{{{}^{-}}^2}\times 8=9.9$ count /sec

  $\begin{array}{l}Thus,totalnumberofcount=4.7+9.9=14.6\hfill \\ Hence,ratioofcountsatt=8\hfill \\ For{}^{198}Au=4.7/14.6=0.32=32\%\hfill \\ Fo{r}^{131}I=9.9/14.6=0.68=68\%\hfill \end{array}$

  $Att=16$

  $\begin{array}{l}{\text{For}}^{198}Au\hfill \\ \lambda =0.257\hfill \\ A_0=36.8\hfill \\ t=16\hfill \end{array}$

  $A=A_0{e}^{-\lambda t}=36.8\times e^{-0.257\times 16}=0.60\text{ count/sec}$

  $\begin{array}{l}For131I,\\ \lambda =8.6\times {10}^{-}2\\ A_0=19.8\\ t=16\end{array}$

  $A=A_0{e}^{-\lambda t}=19.8\times e^{-8.6}\times {10}^{-2}\times 16=5\text{count/sec}$

  $\begin{array}{l}Thus,totalnumberofcount=0.60+5=5.60\hfill \\ Hence,ratioofcountsatt=16\hfill \\ Fo{r}^{198}Au=0.60/5.6=0.11=11\%\hfill \\ {}^{131}Au=5/5.6=0.89=89\%\hfill \end{array}$

Decay activity of a solution is plotted on graph, by plotting count rate per second over a period of 16 days; we get a curve of total activity as a function of time.

The equation of the curve is a summation of activities of two components ${}^{198}Auan{d}^{131}I$ . we have the following data regarding each component:

  $\begin{array}{lll} \hfill & A_0\hfill & \lambda \hfill \\ {}^{198}Au\hfill & 36.8\hfill & 0.257\hfill \\ {}^{131}I\hfill & 19.8\hfill & 0.086\hfill \end{array}$

Equation of curve of activity versus time is:

  $\begin{array}{l}A={\left(A_{0}e{-}^{\lambda t}\right)}_{198Au}+{\left(A_{0}e{-}^{\lambda t}\right)}_{131I}\hfill \\ A=36.8e^{-0.257t}+19.8e^{-0.086t}\hfill \end{array}$

Conclusion:

We calculate the relation count rates over a period of 16 days for ${}^{198}Auan{d}^{131}I$

  $\begin{array}{lll} \hfill & {}^{198}Au\hfill & {}^{131}I\hfill \\ Att=0\hfill & 65\%\hfill & 35\%\hfill \\ Att=3\hfill & 52\%\hfill & 48\%\hfill \\ Att=8\hfill & 32\%\hfill & 68\%\hfill \\ Att=16\hfill & 11\%\hfill & 89\%\hfill \end{array}$

And the corresponding equation of curve obtained by plotting counting rates per second over a period of 16 days is:

  $A=36.8e^{-0.257t}+19.8e^{-0.086t}$

The equation is the addition of activities of ${}^{198}\text{Au and}{}^{131}\text{I}$ .