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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