A) 6.62 years
B) 5 years
C) 3.2 years
D) 7 years
Correct Answer: A
Solution :
\[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/{{T}_{1/2}}}}\Rightarrow {{N}_{A}}=10{{\left( \frac{1}{2} \right)}^{t/1}}\] and \[{{N}_{B}}=1{{\left( \frac{1}{2} \right)}^{t/2}}\] Given \[{{N}_{A}}={{N}_{B}}\Rightarrow 10{{\left( \frac{1}{2} \right)}^{t}}={{\left( \frac{1}{2} \right)}^{t/2}}\] \[\Rightarrow 10={{\left( \frac{1}{2} \right)}^{-t/2}}\Rightarrow 10={{2}^{t/2}}.\]Taking log both the sides. \[{{\log }_{10}}10=\frac{t}{2}{{\log }_{10}}2\Rightarrow 1=\frac{t}{2}\times 0.3010\] \[\Rightarrow t=6.62\]years.
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