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