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Chapter 19: Problem 31

The first atomic explosion was detonated in the desert north of Alamogordo, New Mexico, on July 16, 1945 . What percentage of the strontium- \(90\left(t_{1 / 2}=28.9\right.\) years) originally produced by that explosion still remains as of July 16,2013 ?

Short Answer

Expert verified
As of July 16, 2013, approximately \(19.5\%\) of the strontium-\(90\) originally produced in the first atomic explosion still remains.

Step by step solution

01

Determine the time elapsed

Calculate the time elapsed since the first atomic explosion by subtracting the explosion date from the given date, which is July 16, 2013. Initial date: July 16, 1945 Final date: July 16, 2013 Time elapsed = 2013 - 1945 = 68 years
02

Calculate the number of half-lives

Find how many half-lives have occurred in the elapsed time period. Given half-life (t1/2) of strontium-90: 28.9 years Time elapsed: 68 years Number of half-lives = Time elapsed / Half-life = 68 / 28.9 ≈ 2.354
03

Use the half-life formula

Use the formula for remaining percentage of a substance after n half-lives: Remaining percentage = \((0.5)^n\) * 100%, where n is the number of half-lives. In this case, n ≈ 2.354 Remaining percentage of strontium-90 = \((0.5)^{2.354}\) * 100% ≈ 19.5% So, as of July 16, 2013, approximately 19.5% of the strontium-90 originally produced still remains.

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