A) \[3.6\times {{10}^{10}}\]
B) \[3.6\times {{10}^{12}}\]
C) \[3.1\times {{10}^{15}}\]
D) \[31.1\times {{10}^{15}}\]
Correct Answer: A
Solution :
According to Avogadro's hypothesis. \[{{N}_{0}}=\frac{6.02\times {{10}^{23}}}{226}=2.66\times {{10}^{21}}\] Half-life\[=T=\frac{0.693}{\lambda }=1620\]year \[\therefore \] \[\lambda =\frac{0.693}{1620\times 3.16\times {{10}^{7}}}\] \[=1.35\times {{10}^{-11}}{{s}^{-1}}\] Because half-life is very much large as compared to its times interval, hence\[N={{N}_{0}}\]. Now, \[\frac{dN}{dt}=\lambda N=\lambda {{N}_{0}}\] or \[dN=\lambda {{N}_{0}}dt\] \[=(1.35\times {{10}^{-11}})(2.66\times {{10}^{21}})\times 1\] \[=1.35\times {{10}^{-11}}{{s}^{-1}}\]
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