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The Nucleus — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsThe Nucleus5 Marks
Question
The count rate of nuclear radiation coming from a radiation coming from a radioactive sample containing $^{128}I$ varies with time as follows.
Time t(minute): $0$ $25$ $50$ $75$ $100$
Ctount rate $R(10^9s^{-1}):$ $30$ $16$ $8.0$ $3.8$ $2.0$
  1. Plot In $\Big(\frac{\text{R}_0}{\text{R}}\Big)$ against t.
  2. From the slope of the best straight line through the points, find the decay constant $\lambda.$
  3. Calculate the half-life $\text{t}_{\frac{1}{2}}.$
✓

Answer

  1. Here we should take $R_0$​​​​​​​ at time is $t_0 = 30 \times 10^9s^{-1}​​​​​​​$​​​​​​​
  1. $\text{ln}\Big(\frac{\text{R}_0}{\text{R}_1}\Big)=\text{ln}\Big(\frac{30\times10^9}{30\times10^9}\Big)=0$
  2. $\text{ln}\Big(\frac{\text{R}_0}{\text{R}_2}\Big)=\text{ln}\Big(\frac{30\times10^9}{16\times10^9}\Big)=0.63$
  3. $\text{ln}\Big(\frac{\text{R}_0}{\text{R}_3}\Big)=\text{ln}\Big(\frac{30\times10^9}{8\times10^9}\Big)=1.35$
  4. $\text{ln}\Big(\frac{\text{R}_0}{\text{R}_4}\Big)=\text{ln}\Big(\frac{30\times10^9}{3.8\times10^9}\Big)=2.06$
  5. $\text{ln}\Big(\frac{\text{R}_0}{\text{R}_5}\Big)=\text{ln}\Big(\frac{30\times10^9}{2\times10^9}\Big)=2.7$
  1. $\therefore$ The decay constant $\lambda=0.028\text{min}^{-1}$
  2. $\therefore$ The half life period $\text{t}_{\frac{1}{2}}$
$\text{t}_{\frac{1}{2}}=\frac{0.693}{\lambda}=\frac{0.693}{0.028}=25\text{min}.$

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