A) \[\frac{1}{0.693}h\]
B) \[0.693\times 2h\]
C) \[0.693\times \frac{1}{4}h\]
D) \[0.693\times 8h\]
Correct Answer: A
Solution :
From Rutherford and Soddy law, at any instant the rate of decay of radioactive atoms is proportional to the number of atoms present at that in start that is \[\therefore \] \[N={{N}_{o}}{{e}^{-\lambda t}}\] \[\frac{1}{0.25}={{e}^{-\lambda \times 2}}\] \[\Rightarrow \] \[4={{e}^{-2\lambda }}\] Taking log on both sides, we have \[{{\log }_{e}}4=-2\lambda \] \[\Rightarrow \] \[\lambda =0.693/h\] Also mean-life=\[\frac{1}{decay\,\,cons\tan t\,\left( \lambda \right)}\] \[=\frac{1}{0.693}h\]
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