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 :
\[N={{N}_{0}}{{e}^{-\lambda t}}\] \[\frac{N}{{{N}_{0}}}={{e}^{-\lambda t}}\] \[\frac{1}{0.25}={{e}^{-\lambda t}}\] \[4={{e}^{-2\lambda }}\] \[{{\log }_{e}}4=-2\lambda \] \[\lambda =0.693/h\] Also mean-life =\[\frac{\text{1}}{\text{decay}\,\text{constant( }\!\!\lambda\!\!\text{ )}}\]
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