The answer is 3.125%.
The relation between number of half-lives (n) and amount of sample that remained (x) can be expressed as:
\((1/2)^{n} =x\)
It is given:
n = 5
To find x, we will change n in the equation with five:
\((1/2)^{n} =x\)
⇒ \((1/2)^{5} =x\)
⇒ \(0.5^{5} =x\)
⇒ \(x=0.03125 = \frac{3.125}{100} = 3.125%\)
Therefore, 3.125% of a radioactive species would be found as daughter material after five half-lives.
