Case study (4 Marks)

Question 14 Marks

The half-life of ${ }^{226} \mathrm{Ra}$ is 1602 y . Calculate the activity of 0.1 g of $\mathrm{RaCl}_2$ in which all the radium is in the form of ${ }^{226} \mathrm{Ra}$. Taken atomic weight of Ra to be $226 \mathrm{~g} / \mathrm{mol}^{-1}$ and that of Cl to be $35.5 \mathrm{~g} / \mathrm{mol}^{-1}$.

Answer

$\text{t}_{\frac{1}{2}}=1602\text{Y};\text{ Ra}=226\text{g/mole};\text{ Cl}=35.5\text{g/mole}.$
1 mole $RaCl_2 = 226 + 71 = 297g$
297g = 1 mole of Ra.
$0.1\text{g}=\frac{1}{297}\times0.1\text{ mole of Ra}=\frac{0.6\times6.023\times10^{23}}{297}\\=0.02027\times10^{22}$
$\lambda=\frac{0.693}{\text{t}_{\frac{1}{2}}}=1.371\times10^{-11}$
Activity $\lambda\text{N}=1.371\times10^{-11}\times2.027\times10^{20}$
$=2.779\times10^{9}=2.8\times10^9$ disintegrations/second.

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Question 24 Marks

A sample contains a mixture of $^{108}Ag$ and $^{110}Ag$ isotopes each having an activity of 8.0 \times 10^8 disintegration per second. $^{110}Ag$ is known to have larger half-life than $^{108}Ag.$ The activity A is measured as a function of time and the following data are obtained.

Time (s)	Activity (A) $(10^8$ disinte- grations $s^{-1})$	Time (s)	Activity (A) $(10^8$ disinte-grations $s^{-1})$
20	11.799	200	3.0828
40	9.1680	300	1.8899
60	7.4492	400	1.1671
80	6.2684	500	0.7212
100	5.4115

Plot ln $\Big(\frac{\text{A}}{\text{A}_0}\Big)$ versus time.
See that for large values of time, the plot is nearly linear. Deduce the half-life of $^{110}Ag$ from this portion of the plot.
Use the half-life of $^{110}Ag$ to calculate the activity corresponding to $^{108}Ag$ in the first 50s.
Plot In $\Big(\frac{\text{A}}{\text{A}_0}\Big)$ versus time for $^{108}Ag$ for the first 50s.
Find the half-life of $^{108}Ag.$

Answer

Activities of sample containing $^{108}Ag$ and $^{110}Ag$ isotopes $= 8.0 \times 10^8$ disintegration/sec.

Here we take $A = 8 × 108$ dis./sec

$\text{ln}\Big(\frac{\text{A}_1}{\text{A}_{0_{1}}}\Big)=\text{ln}\Big(\frac{11.79}{8}\Big)=0.389$
$\text{ln}\Big(\frac{\text{A}_2}{\text{A}_{0_{2}}}\Big)=\text{ln}\Big(\frac{9.1680}{8}\Big)=0.1362$
$\text{ln}\Big(\frac{\text{A}_3}{\text{A}_{0_{3}}}\Big)=\text{ln}\Big(\frac{7.4492}{8}\Big)=-0.072$
$\text{ln}\Big(\frac{\text{A}_4}{\text{A}_{0_{4}}}\Big)=\text{ln}\Big(\frac{6.2684}{8}\Big)=-0.244$
$\text{ln}\Big(\frac{5.4115}{8}\Big)=-0.391$
$\text{ln}\Big(\frac{3.0828}{8}\Big)=-0.954$
$\text{ln}\Big(\frac{1.8899}{8}\Big)=-1.443$
$\text{ln}\Big(\frac{1.167}{8}\Big)=-1.93$
$\text{ln}\Big(\frac{0.7212}{8}\Big)=-2.406$

The half life of $^{110}Ag$ from this part of the plot is 24.4s.
Half life of $^{110}Ag = 24.4s.$

$\therefore\text{decay constant}\lambda=\frac{0.693}{24.4}=0.0284\Rightarrow\text{t}=50\text{sec,}$
The activity $\text{A = A}_{0}\text{e}^{-\lambda\text{t}}=8\times10^8\times\text{e}^{-0.0284\times50}=1.93\times10^8$

The half life period of $^{108}Ag$ from the graph is 144s.

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Physics Case study (4 Marks)

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