Question

A bone fragment has 67% of the parent Pb-210 remaining. The half-life of Pb-210 is 22 years. How old is the bone fragment?

Answer 1

To solve this question we will use two important formulas.

The first formula involves the calculation of the decay constant, $k$ which is to be found as:

$k=\frac{0.6931}{T_{\frac{1}{2}}}$

where $T_{\frac{1}{2}}$ is the half life of Pb-210 which is given to be 22 years in the question.

Thus, $k=\frac{0.6931}{22}\approx0.0315$

Now, we will use the second formula which is called the function of decay formula and is given as:

$N = N_{0} e^{-kt}$

where $N$ percentage of Pb-210 remaining in the bone at present, which in our case is 67% or 0.67

$N_{0}$ percentage of Pb-210 to start with, which is always 100% or 1.

Plugging in all these values in the decay formula we are supposed to find the time, $t$.

$0.67=1\times e^{-0.0315t}$

Taking natural log on both sides we get:

$ln(0.67)=-0.0315t$

$\approx-0.4005=-0.0315t$

$\therefore t=\frac{0.4005}{0.0315}$

$t\approx12.71$ years

Therefore the bone fragment is about 12.71 years old.