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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