A) 2.5 g
B) 5 g
C) 10 g
D) 15 g
Correct Answer: B
Solution :
After\[n-\]half lines quantity of radiatioactive substance left is \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/{{T}_{1/2}}}}\] Given, \[{{T}_{1/2}}=140\,days,t=28\,days,\,{{N}_{0}}=20g\] \[\therefore \] \[N=20{{\left( \frac{1}{2} \right)}^{\frac{280}{140}}}=\frac{20}{4}=5\,g\]
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