A) \[0.0019\,\,mg\]
B) \[1.019\,\,mg\]
C) \[1.109\,\,mg\]
D) \[0.019\,\,mg\]
Correct Answer: D
Solution :
Half-life\[{{T}_{1/2}}=3.6\,\,days\] Initial quantity\[{{N}_{0}}=20\,\,mg\] Total line\[=36\,\,days\] The number of half lives \[n=\frac{t}{{{T}_{1/2}}}=\frac{36}{3.6}=10\] Hence, mass of radioactive substance left after 10 half lives \[N={{N}_{0}}\times {{\left( \frac{1}{2} \right)}^{n}}\] \[=20\times \frac{1}{1024}=0.019\,\,mg\]
You need to login to perform this action.
You will be redirected in
3 sec
