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Chapter 19: Problem 29

The first atomic explosion was detonated in the desert north of Alamogordo, New Mexico, on July 16, 1945. What fraction of the strontium- \(90\left(t_{1 / 2}=28.9\right.\) years) originally produced by that explosion still remains as of July \(16,2009 ?\)

Short Answer

Expert verified
As of July 16, 2009, approximately \(6.749\%\) of the strontium-90 originally produced by the first atomic explosion still remains.

Step by step solution

01

Calculate the time elapsed

To find the elapsed time, subtract the initial year from the final year: Elapsed time (t) = 2009 - 1945 t = 64 years
02

Apply the half-life formula

Now, we will use the half-life formula to find the remaining fraction of strontium-90: \(N_t = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}\) We are interested in finding the remaining fraction of strontium-90 (\(\frac{N_t}{N_0}\)), so we can rearrange the equation: \(\frac{N_t}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}\) Plugging in the values we have: \(\frac{N_t}{N_0} = \left(\frac{1}{2}\right)^{\frac{64}{28.9}}\)
03

Calculate the remaining fraction

Using a calculator, compute the remaining fraction of strontium-90: \(\frac{N_t}{N_0} \approx 0.06749\)
04

Express the fraction as a percentage

To express the remaining fraction as a percentage, multiply the fraction by 100: Percentage remaining = 0.06749 × 100 = 6.749% As of July 16, 2009, approximately 6.749% of the strontium-90 originally produced by the first atomic explosion still remains.

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