Question

potassium - 40 has a half - life of 1.277 × 10^9 years. after 1.022 × 10^10 years, how much potassium - 40 will remain from a 500.3 - g sample?
approximately 3.91 g
approximately 62.54 g
approximately 71.47 g
approximately 1.95 g

Explanation:

Step1: Calculate the number of half - lives

The formula for the number of half - lives \(n=\frac{t}{T}\), where \(t = 1.022\times10^{10}\) years and \(T=1.277\times 10^{9}\) years.

$$n=\frac{1.022\times 10^{10}}{1.277\times 10^{9}}=\frac{10.22}{1.277}\approx8$$

Step2: Use the radioactive decay formula

The radioactive decay formula is \(N = N_{0}\times(\frac{1}{2})^{n}\), where \(N_{0}=500.3\) g and \(n = 8\)

$$N=500.3\times(\frac{1}{2})^{8}$$

$$N = 500.3\times\frac{1}{256}$$

$$N=\frac{500.3}{256}\approx1.95$$

Answer:

\(1.95\) g