# Question #c4d7e

Dec 29, 2014

A quick way to solve nuclear half-life problems is by remembering that, esentially, you are dividing whatever amount you have by 2 for every half-life.

So, if you start with, say, 100 g of a radioactive isotope that has a half-life of 1 million years, you will have

$\frac{100}{2} = 50$ $\text{g}$ after the first 1 million years,

$\frac{50}{2} = 25$ $\text{g}$ after another 1 million years,

$\frac{25}{2} = 12.5$ $\text{g}$ after another 1 million years,

$\frac{12.5}{2} = 6.25$ $\text{g}$ after another 1 million years, and so on...

So, for this example, you are left with 1/4th of the original sample after 2 "cycles". Therefore, since, in your case, the element's half-life is 703 million, 2 "cycles" mean

703 + 703 = 1406 million = 1.4 billion years.

If the problem would have asked for, say, 1/16th of the original sample, that would have corresponded to

$\frac{1}{2} ^ \left(\text{number of cycles}\right) = \frac{1}{16}$, so you'd have

${2}^{\text{number of cycles}} = 16 \to$ 4 cycles. Therefore,

$4 \cdot 703$ $\text{million} = 2812$ $\text{million} = 2.8$ $\text{billion years}$ need to pass to reach 1/16th of the original sample.