A) \[3.61\times {{10}^{10}}\]
B) \[3.6\times {{10}^{12}}\]
C) \[3.1\times {{10}^{15}}\]
D) \[31.1\times {{10}^{15}}\]
Correct Answer: A
Solution :
Rate of disintegration\[\frac{dN}{dt}=\lambda N\] From Avogadros principle Number of atoms in 1 g radium \[=\frac{6\times {{10}^{23}}}{226}\] Decay constant\[\lambda =\frac{1}{\tau }\] \[=\frac{0.693}{1620}yr\] \[=\frac{0.693}{1620\times 365\times 24\times 60\times 60}\] \[=\frac{0.693}{1620\times 3.15\times {{10}^{7}}}{{s}^{-1}}\] \[\therefore \]\[\frac{dN}{dt}=\frac{0.693}{1620\times 3.15\times {{10}^{7}}}\times \frac{6\times {{10}^{23}}}{226}\] \[=3.61\times {{10}^{10}}\,\text{per}\,\,\text{sec}\]You need to login to perform this action.
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