4 Marks Questions

Question 14 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 $A = 8 \times 108 dis./sec$

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

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 $144s.$

View full question & answer→

Question 24 Marks

The half-life of $^{226}Ra$ is $1602y.$ Calculate the activity of $0.1g$ of $RaCl_2$ in which all the radium is in the form of $^{226}Ra.$ Taken atomic weight of $Ra$ to be $226\ g/mol^{-1}$ and that of $Cl$ to be $35.5\ g/mol^{-1}.$

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.

View full question & answer→

Physics 4 Marks Questions

Test yourself on this topic