A) 16.5 days
B) 3.8 days
C) 33 days
D) 76 days
Correct Answer: A
Solution :
From decay equation \[N={{N}_{0}}{{e}^{-\lambda t}}\] \[\lambda =\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{3.8}\] \[N=\frac{{{N}_{0}}}{20}\] \[\therefore \] \[\frac{{{N}_{0}}}{20}={{N}_{0}}{{e}^{-\frac{0.693}{3.8}t}}\] \[\frac{1}{20}={{e}^{-\frac{0.693}{3.8}t}}\] \[{{e}^{\frac{0.693}{3.8}t}}=20\] \[\log 20=\frac{0.693}{3.8}t\,{{\log }_{10}}e\] \[1.3010=\frac{0.693}{3.8}\times t\times 0.4343\] \[t=16.5day\]
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