A) 100
B) 10
C) 1
D) 0.1
Correct Answer: B
Solution :
\[k=\frac{2.303}{t}\log \frac{{{N}_{0}}}{N}\] \[=\frac{2.303}{23.03}\log \frac{2\times {{10}^{7}}}{2\times {{10}^{6}}}\] \[=\frac{2.303}{23.03}\log 10\] \[k=0.1\text{ }mi{{n}^{-1}}\] \[{{t}_{1/2}}=\frac{0.693}{k}=\frac{0.693}{0.1}=0.93\,\min \] average life \[(T)=1.44\times {{t}_{1/2}}\] \[=1.44\times 6.93=10\text{ }min\]
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