A) 1.042 g
B) 2.084 g
C) 3.125 g
D) 4.167 g
Correct Answer: C
Solution :
From radioactive decay equation \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/{{t}_{1/2}}}}\] \[\therefore \] \[N=200\,\,{{\left( \frac{1}{2} \right)}^{24/2}}\] \[=200\,\,{{\left( \frac{1}{2} \right)}^{6}}\] \[=\frac{200}{64}=3.125\,g\]
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