Question

The half life of Pb-210 is 22 years. A decayed animal shows 25% of the original Pb-210 remains; how long has the animal been deceased to the nearest tenth of a year?

Answer 1

This is an easy half-life calculation. (Most of these are NOT easy).
If 25% remains of an original amount of 100%, then the time elapsed is precisely 44 years. (2 half-lives)
In 22 years (one half-life), it would be 50%
In 44 years (two half-lives), it would be 25%
In 66 years (three half-lives), it would be 12.5%
In 88 years (four half-lives), it would be 6.25%

Answer 2

44 years.

Explanation:

Let original amount be 100.

We have been given that the half life of Pb-210 is 22 years. A decayed animal shows 25% of the original Pb-210 remains.

We will use half-life formula to solve our given problem.

`A=a\cdot (0.5)^{(t)/(h)}`, where,

A = Amount after t units of time,

a = Initial amount,

t = Time,

h = Half-life.

`25=100\cdot (0.5)^{(t)/(22)}`

`(25)/(100)=\frac{100\cdot (0.5)^{(t)/(22)}}{100}`

`0.25=(0.5)^{(t)/(22)}`

Take natural log of both sides:

`\text{ln}(0.25)=\text{ln}((0.5)^{(t)/(22)})`

Using natural log property
`\text{ln}(a^b)=b\cdot \text{ln}(a)`, we will get:

`\text{ln}(0.25)=(t)/(22)*\text{ln}(0.5)`

`\frac{\text{ln}(0.25)}{\text{ln}(0.5)}=\frac{t*\text{ln}(0.5)}{22*\text{ln}(0.5)}`

`(-1.38629436)/(-0.69314718)=(t)/(22)`

`2=(t)/(22)`

`2*22=(t)/(22)*22`

`44=t`

Therefore, the animal has been deceased to 44 years.