A) \[2.5\,mCi\]
B) \[1\,mCi\]
C) \[3.2\,mCi\]
D) \[3\,mCi\]
Correct Answer: B
Solution :
As, \[\frac{{{N}_{32}}}{{{N}_{33}}}=\frac{4}{1}\] \[[{{N}_{33}}={{N}_{0}}]\] \[\Rightarrow \] \[\frac{dN}{N}={{\lambda }_{1}}{{N}_{32}}+{{\lambda }_{2}}{{N}_{33}}\] \[\Rightarrow \] \[7\times {{10}^{-3}}={{\lambda }_{1}}4{{N}_{0}}+{{\lambda }_{2}}{{N}_{0}}\] \[\Rightarrow \] \[7\times {{10}^{-3}}={{N}_{0}}\,\left[ \frac{In\,(2)}{5}+\frac{In\,(2)}{30} \right]\] \[\Rightarrow \] \[{{N}_{0}}=\frac{30\times {{10}^{-3}}}{In\,(2)}\] After 60 days, \[\frac{dN}{dt}={{\lambda }_{1}}\frac{4{{N}_{0}}}{8}+\frac{{{\lambda }_{2}}{{N}_{0}}}{4}=1\,mCi\]You need to login to perform this action.
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