A) \[3.61\times {{10}^{10}}\]
B) \[3.6\times {{10}^{12}}\]
C) \[3.11\times {{10}^{15}}\]
D) \[31.1\times {{10}^{15}}\] (Avogadro's number \[N=6.02\times {{10}^{26}}\]atom/kilomol)
Correct Answer: A
Solution :
\[\frac{dN}{dt}=\lambda N;\] \[\lambda =\frac{0.6931}{{{t}_{12}}}=\frac{0.6931}{1620\times 365\times 24\times 60\times 60},\] \[N=\frac{6.023\times {{10}^{23}}}{226}\] \[\therefore \frac{dN}{dt}=\frac{0.6931\times 6.023\times {{10}^{23}}}{1620\times 365\times 24\times 60\times 60\times 226}\]\[\frac{1}{2}M{{v}^{2}}=qV.\]
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