A) \[3.61\,\times {{10}^{10}}\]
B) \[3.6\,\times {{10}^{12}}\]
C) \[3.11\,\times {{10}^{15}}\]
D) \[31.\,1\times {{10}^{15}}\]
Correct Answer: A
Solution :
From the formula \[\frac{dN}{dt}=\lambda N\] ?(i) and \[\lambda =\frac{0.693}{{{T}_{1/2}}}\] \[=\frac{0.693}{1620\times 365\times 24\times 60\times 60}\] ?(ii) and \[N=\frac{6.023\times {{10}^{23}}}{226}\] ?(iii) Now from equations (ii) and (iii), putting the values of K and N in equation (i), we get \[\frac{dN}{dt}=\frac{0.693\times 6.023\times {{10}^{23}}}{1620\times 365\times 24\times 60\times 226}\] \[=3.61\times {{10}^{10}}\]
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