The answer is 3.
The relation between number of half-lives (n) and decimal amount remaining (x) can be expressed as:
\((1/2) ^{n} =x\)
We need to calculate n, but we need x to do that. To calculate what percentage of a radioactive species would be found as daughter material, we must calculate what amount remained:
1.28 - 1.12 = 0.16
If 1.28 is 100%, how much percent is 0.16:
1.28 : 100% = 0.16 : x
x = 12.5%
Presented as decimal amount:
x = 0.125
Now, let's implement this in the equation:
\((1/2) ^{n} =0.125\)
Because of the exponent, we will log both sides of the equation:
\(n * log(1/2) = log(0.125)\)
\(n = \frac{log(0.125)}{log(1/2)}\)
\(n = \frac{log(0.125)}{log(0.5)}\)
\(n= \frac{-0.903}{-0.301}\)
\(n = 3\)
Therefore, 3 half-lives have passed since the sample originally formed.
