# A sample of U-238 initially contained 1.3x10^18 atoms when the universe was formed 13.8 billion years ago, how many U-238 atoms will it contain today?

Jul 12, 2016

$N \left(t\right) = 1.5 \times {10}^{17}$

#### Explanation:

We can find the number of atoms remaining from an original sample using the half-life formula:

$N \left(t\right) = N \left(0\right) {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(1 / 2\right)}$

where
${t}_{1 / 2}$ is the half-life of U-238 which is $4.468 \times {10}^{9}$ years ($4.468$ billion years) and $N \left(0\right)$ is the original number of atoms in the sample

Plugging in our numbers we get:

$N \left(t\right) = 1.3 \times {10}^{18} {\left(\frac{1}{2}\right)}^{\frac{13.8 \times {10}^{9}}{4.468 \times {10}^{9}}} = 1.5 \times {10}^{17}$

Checking to see if this makes sense, we are left with about 12% of the original sample and our time was approximately $3$ half-lives. Exactly 3 half-lives would be:

1/2*1/2*1/2 = 1/8 = 0.125 = 12.5%

Therefore our answer does make sense.