A) \[5.24\times {{10}^{9}}\] disintegrations/s
B) \[5.24\times {{10}^{10}}\] disintegrations/s
C) \[5.24\times {{10}^{8}}\] disintegrations/s
D) \[5.24\times {{10}^{11}}\] disintegrations/s
Correct Answer: A
Solution :
The instant activity of radioactive isotope is \[R=\lambda N\] ...(i) where\[\lambda \]is decay constant for isotope and N the number of atoms. Also, \[\lambda =\frac{0.693}{T}\] ...(ii) \[\therefore \]From Eqs. (i) and (ii), we have \[R=\frac{0.693}{T}N\] Given, \[T=28\text{ }yr=2.8\times 3.65\times 2.4\times 3.6\text{ }s\] \[N=\frac{6.02\times {{10}^{23}}}{9}\] \[\therefore \] \[R=\frac{0.693\times 6.02}{2.8\times 3.65\times 2.4\times 3.6\times 9}\times {{10}^{20-8}}\] \[\Rightarrow \]\[R=5.24\times {{10}^{12-3}}=5.24\times {{10}^{9}}disint/s\].
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