Question

The half-life of cesium-137 is 30 years. Suppose that we start with 60 grams of cesium-137 in a storage pool.

How many half-lives will it take for there to be 10 grams of cesium-137 in the storage pool? (Round answer to two decimal places.)

How many years is that? (Round answer to one decimal place.)

Answer 1

9514 1404 393

Answer:

• 2.58 half-lives
• 77.5 years

Explanation:

The remaining amount, in terms of t half-lives, is ...

  q(t) = 60·(1/2)^t

We want t when q(t) = 10, so ...

  10 = 60·(1/2)^t

  1/6 = (1/2)^t . . . . . . divide by 60

  log(1/6) = t·log(1/2) . . . . take logs

  t = log(1/6)/log(1/2) = -0.778151/-0.30103 ≈ 2.58496

It will take about 2.58 half-lives for there to be 10 grams remaining.

In years, that is 2.58×30 = 77.5 years.