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