Probability of a disintegration

Question

The half-life of Uranium-$238$ is $5×10^9$ years. What is the probability than a uranium atom disintegrates in any one year?

I think I have to use Poisson's law but I don't know how to apply it in this context.

Thank you in advance for any hints!

Answer 1

The decay rate $\lambda$ is closely related to half-life, $$ \lambda = \frac{\ln (2)}{t_{1/2}} $$

This gives the value you are looking for directly, $$\frac{\ln (2)}{5\cdot 10^9} = 1.39\cdot 10^{-10}$$

This is the instantaneous rate of change in the amount of uranium. The total amount of uranium is reducing at a rate of $1.39\cdot 10^{-10}$ per year. Therefore - since the change in this number is negligible over the course of a year, with such a long half-life - this is the probability associated with each atom in the uranium that it will decay over that year.

At the end of the year, the probabilities will have played out for each one of the atoms, and there will be $(1-1.39\cdot 10^{-10})$ times the initial amount of uranium left at the end of the year.

Answer 2

Well after more thinking there is a "best" demonstration, because of the geometric law is without memory we search the probability $p$ for which :

$$ (1-p)^{5 \cdot 10^9}=\frac{1}{2} $$

After calculs we have :

$$ p=\frac{\ln(2)}{5 \cdot 10^9} \approx 1.4 \cdot 10^{-10} $$

Shadock ! :D