A) 10 min
B) 20 min
C) 30 min
D) 40 min
Correct Answer: B
Solution :
\[\lambda =\frac{0.693}{{{T}_{1/2}}}=\frac{0.693}{20}=0.03465\] Now time of decay \[t=\frac{2.303}{\lambda }\log \frac{{{N}_{0}}}{N}\] \[\Rightarrow {{t}_{1}}=\frac{2.303}{0.03465}\log \frac{100}{67}=11.6\min \]min and \[{{t}_{2}}=\frac{2.303}{0.03465}\log \frac{100}{33}=32min\] Thus time difference between points of time = t1 ? t2 =32 ? 11.6 = 20.4 min \[\approx \] 20 min.
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