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The Nucleus — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsThe Nucleus5 Marks
Question
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    
  1. Plot ln $\Big(\frac{\text{A}}{\text{A}_0}\Big)$ versus time.
  2. 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.
  3. Use the half-life of $^{110}Ag$ to calculate the activity corresponding to $^{108}Ag$ in the first 50s.
  4. Plot In $\Big(\frac{\text{A}}{\text{A}_0}\Big)$ versus time for $^{108}Ag$ for the first 50s.
  5. 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.
  1. Here we take A = 8 × 108 dis./sec
  1. ​​​​​$\text{ln}\Big(\frac{\text{A}_1}{\text{A}_{0_{1}}}\Big)=\text{ln}\Big(\frac{11.79}{8}\Big)=0.389$
  2. $\text{ln}\Big(\frac{\text{A}_2}{\text{A}_{0_{2}}}\Big)=\text{ln}\Big(\frac{9.1680}{8}\Big)=0.1362$
  3. $\text{ln}\Big(\frac{\text{A}_3}{\text{A}_{0_{3}}}\Big)=\text{ln}\Big(\frac{7.4492}{8}\Big)=-0.072$
  4. $\text{ln}\Big(\frac{\text{A}_4}{\text{A}_{0_{4}}}\Big)=\text{ln}\Big(\frac{6.2684}{8}\Big)=-0.244$
  5. $\text{ln}\Big(\frac{5.4115}{8}\Big)=-0.391$
  6. $\text{ln}\Big(\frac{3.0828}{8}\Big)=-0.954$
  7. $\text{ln}\Big(\frac{1.8899}{8}\Big)=-1.443$
  8. $\text{ln}\Big(\frac{1.167}{8}\Big)=-1.93$
  9. $\text{ln}\Big(\frac{0.7212}{8}\Big)=-2.406$
  1. The half life of $^{110}Ag$ from this part of the plot is 24.4s.
  2. 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 $\text{A = A}_{0}\text{e}^{-\lambda\text{t}}=8\times10^8\times\text{e}^{-0.0284\times50}=1.93\times10^8$
  1.  
  1. The half life period of $^{108}Ag$ from the graph is 144s.

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