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What is the half-life of bismuth-210 if 15.6 days are required for the activity of a sample of bismuth-210 to fall to 11.6 percent of its...

I suppose there are no other radioactive elements in the sample, and that the results of bismuth-210's decay are not radioactive. Then the activity level is proportional to the remaining quantity of bismuth-210.


Then denote the half-life of bismuth-210 as `H` days (we select days as the units of measure for convenience, because the given time is in days). Then after `H` days a half of the initial activity remains, after `2H` days a quarter of initial activity remains and after `x` days `1/2^(x/H) = 2^(-x/H)` remains. This is true for any `x.`


It is given that for `x=15.6` days 11.6% = 0.116-th part remains, so


`2^(-15.6/H) = 0.116.`



Take the base `2` logarithm of both sides and obtain


`-15.6/H = log_2(0.116),` or `H = -15.6/log_2(0.116)=-15.6 (ln(2))/(ln(0.116)).`


Here we used that `log_2(a) = (ln(a))/(ln(2)).`


In numbers it is about  5 (days).

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