4 Marks Question

Question 14 Marks

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

Time (s)	Activity (A) (10⁸ disinte- grations s^-1)	Time (s)	Activity (A) (10⁸ 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 ¹¹⁰Ag from this portion of the plot.
Use the half-life of ¹¹⁰Ag to calculate the activity corresponding to ¹⁰⁸Ag in the first 50s.
Plot In $\Big(\frac{\text{A}}{\text{A}_0}\Big)$ versus time for ¹⁰⁸Ag for the first 50s.
Find the half-life of ¹⁰⁸Ag.

Answer

Activities of sample containing ¹⁰⁸Ag and ¹¹⁰Ag isotopes = 8.0 × 10⁸ 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 ¹¹⁰Ag from this part of the plot is 24.4s.
Half life of ¹¹⁰Ag = 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 ¹⁰⁸Ag from the graph is 144s.

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Question 24 Marks

The half-life of ²²⁶Ra is 1602y. Calculate the activity of 0.1g of RaCl₂ in which all the radium is in the form of ²²⁶Ra. Taken atomic weight of Ra to be 226g/mol^-1 and that of Cl to be 35.5g/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 RaCl₂ = 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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