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_{(1)/(2)}}`

where
`T_{(1)/(2)}` is the half life of Pb-210 which is given to be 22 years in the question.

Thus,
`k=(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* e^(-0.0315t)`

Taking natural log on both sides we get:

`ln(0.67)=-0.0315t`

`\approx-0.4005=-0.0315t`

`\therefore t=(0.4005)/(0.0315)`

`t\approx12.71` years

Therefore the bone fragment is about 12.71 years old.