Question

Question #c4d7e

Answer 1

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

`100/2 = 50` `"g"` after the first 1 million years,

`50/2 = 25` `"g"` after another 1 million years,

`25/2 = 12.5` `"g"` after another 1 million years,

`12.5/2 = 6.25` `"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

`1/(2)^("number of cycles") = 1/16`, so you'd have

`2^("number of cycles") = 16 ->` 4 cycles. Therefore,

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