Question

Question #cf178

Answer 1

It will never lose ALL of it.

Explanation:

After 1590 years it will have lost 50%
After 2x1590 years only 25% will be left
After 3x1590 years 12.5% will be left.
Etc.

Even after 10x1590=15900 years about 0.1% will be left.

The activity can be represented in a graph (NOT to scale):
graph{10*0.7^x [-4.02, 24.47, -4.12, 10.12]}

Answer 2

I got a finite value of `t` in years.

Explanation:

If we begin with `A_0` number of atoms of a sample of radioactive material, number of atoms remaining after time `t` is given by the expression

`A(t)=A_0e^(-0.693t/T_(1//2))`
where `T_(1//2)` is half life of the sample

For `color (white).^226"Ra"` whose half life is given the equation reduces to
`A(t)=A_0e^(-0.693t/1590)`
`=>A(t)=A_0e^(-0.000462t)`

Number of atoms in earth are estimated as `=10^50`.
The isotope of radium which is being considered occurs in traces.
If we start with `10^30` number of atoms of radium today, and put the final remaining number as `1`, as last atom of radium can not decay in to its fraction, we get

`1=10^30xxe^(-0.000462t)`
`=>e^(-0.000462t)=10^(-30)`
Taking `ln` of both sides and solving for `t`

`t~~150,000 " "years`
we get a finite value of `t` in years.