A) \[0.04\,\mu C\]
B) \[0.08\,\mu C\]
C) \[0.24\,\mu C\]
D) \[0.16\,\mu C\]
Correct Answer: D
Solution :
| The activity of a radioactive substance is |
| \[R={{R}_{0}}\,{{\left( \frac{1}{2} \right)}^{n}}\] |
| Here, \[n=\] number of half-lives |
| \[=\frac{t}{{{T}_{1/2}}}=\frac{24}{6}=4\] |
| \[R=0.01\,\mu C\] |
| Hence, |
| \[0.01={{R}_{0}}{{\left( \frac{1}{2} \right)}^{4}}\] |
| or \[{{R}_{0}}=0.01\times {{(2)}^{4}}\] |
| \[=0.01\times 16=0.16\,\mu C\] |
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