A) 4 mg
B) 8 mg
C) 1 mg
D) 2 mg
Correct Answer: D
Solution :
\[{{t}_{1/2}}=12.3\,y.\] Initial amount \[({{N}_{0}})=32\,mg\] Amount left (N) = ? Total time (T) = 49.2 y \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}\] where n = total number of half-life \[n=\frac{Total\,time}{Half\,-\,life}\] \[\frac{49.2}{12.3}=4\] \[So,\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}\] \[\frac{N}{32}={{\left( \frac{1}{2} \right)}^{4}}\] \[\frac{N}{32}=\frac{1}{16}\] \[N=\frac{32}{16}\,=2mg\]You need to login to perform this action.
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