# Question #cf178

May 5, 2017

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]}

May 18, 2017

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

$A \left(t\right) = {A}_{0} {e}^{- 0.693 \frac{t}{T} _ \left(1 / 2\right)}$
where ${T}_{1 / 2}$ is half life of the sample

For ${\textcolor{w h i t e}{.}}^{226} \text{Ra}$ whose half life is given the equation reduces to
$A \left(t\right) = {A}_{0} {e}^{- 0.693 \frac{t}{1590}}$
$\implies A \left(t\right) = {A}_{0} {e}^{- 0.000462 t}$

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

$1 = {10}^{30} \times {e}^{- 0.000462 t}$
$\implies {e}^{- 0.000462 t} = {10}^{- 30}$
Taking $\ln$ of both sides and solving for $t$

$t \approx 150 , 000 \text{ } y e a r s$
we get a finite value of $t$ in years.