A) \[50\text{ }min\]
B) \[100\text{ }min\]
C) \[99\text{ }min\]
D) \[150\text{ }min\]
Correct Answer: C
Solution :
Radioactive disintegration is a 1st order reaction, hence \[k=\frac{2.303}{t}\log \frac{a}{a-x}\] \[\therefore \] \[k=\frac{2.303}{15}\log \frac{100}{100-50}\] also \[k=\frac{2.303}{t}\log \frac{100}{100-99}\] \[\therefore \] \[\frac{2.303}{15}\log \frac{100}{50}=\frac{2.303}{t}\log \frac{100}{1}\] \[\frac{1}{15}\log 2=\frac{1}{t}\log 100\] \[\frac{0.3010}{15}=\frac{2}{t}\] \[t=\frac{2\times 15}{0.3010}=99\min \]
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