Time required to complete 99% decay is k times the time required to complete 90% decay. What is k?

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Solution

For 99% decay, $t = \frac{2.303}{k} l o g \frac{100}{1} = 4.6 / k$
For 90% decay, $t^{'} = \frac{2.303}{k} l o g \frac{100}{10} = 2.3 / k$
Thus, time required to complete 99% decay is how much to time required to complete 90% decay.

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