A) 1.1 times that of half-life
B) 2.2 times that of half-life
C) 3.3 times that of half-life
D) 4.4 times that of half-life
Correct Answer: C
Solution :
[c] 3.3 times that of half-life \[{{t}_{90%}}=\frac{2.303}{k}\,\log \frac{a}{a-0.9a}\] \[=\frac{2.303}{k}\log \,10\] \[=\frac{2.303}{k}\] \[{{t}_{{}^{1}/{}_{2}}}=\frac{2.303}{k}\log \frac{a}{a-{}^{a}/{}_{2}}\] \[{{t}_{{}^{1}/{}_{2}}}=\frac{2.303}{k}\log \frac{2a}{a}\] \[=\frac{2.303\times 0.3010}{k}\] \[\therefore \,\,\frac{{{t}_{90%}}}{{{t}_{{}^{1}/{}_{2}}}}=\frac{1}{0.3010}=3.3\Rightarrow {{t}_{90%}}={{t}_{{}^{1}/{}_{2}}}\times 3.3\]
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