A) 10 min
B) 20 min
C) 30 min
D) 40 min
Correct Answer: C
Solution :
After n half-lives the quantity of a radioactive substance left intact (undecayed) is given by \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] \[={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/{{T}_{1/2}}}}\] Here \[N=\frac{1}{16}{{N}_{0}},\,T=2h\] \[\frac{1}{16}{{N}_{0}}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{2/{{T}_{1/2}}}}\] \[{{\left( \frac{1}{2} \right)}^{4}}={{\left( \frac{1}{2} \right)}^{2/{{T}_{1/2}}}}\] Equating the powers on both sides \[4=\frac{2}{{{T}_{1/2}}}\] \[{{T}_{1/2}}=\frac{1}{2}h=30\min \]
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