A) \[2.5\,\,g\]
B) \[5\,\,g\]
C) \[10\text{ }g\]
D) \[15\text{ }g\]
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, \[{{T}_{1/2}}=140\,\,days,\,\,t=28\,\,days,\,\,\,{{N}_{0}}=20g\] \[\therefore \] \[N=20{{\left( \frac{1}{2} \right)}^{\frac{280}{140}}}=\frac{20}{4}=5g\]
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