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Answer 1 · 2014-12-29T20:52:34

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

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

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

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

$\frac{12.5}{2} = 6.25$ $12.5/2 = 6.25$ $g$ $"g"$ 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

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

$2^{number of cycles} = 16 \to$ $2^("number of cycles") = 16 ->$ 4 cycles. Therefore,

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

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