A) 12.1 days
B) 24 days
C) 60 days
D) 4 days
E) 14 days
Correct Answer: A
Solution :
The decay constant is \[\lambda =\frac{0.693}{T}=\frac{0.693}{3.64}=0.1904\text{/}day\] Also, \[N={{N}_{0}}{{e}^{-\lambda t}}\] Given, \[\frac{N}{{{N}_{0}}}=0.1={{10}^{-1}}\] \[\therefore \] \[{{10}^{-1}}={{e}^{-\lambda t}}\] \[\Rightarrow \] \[{{e}^{\lambda t}}=10\] \[\lambda t={{\log }_{e}}10\] \[=2.3026\times {{\log }_{10}}10\] \[=2.3026\times 1\] \[\therefore \] \[t=\frac{2.3026\times 1}{\lambda }=\frac{2.3026\times 1}{0.1904}=12.1\,days\]
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