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 1.95 g approximately 3.91 g approximately 62.54 g approximately 71.47 g

Explanation:

Step1: Calculate the number of half - lives

The formula for the number of half - lives $n=\frac{t}{T_{1/2}}$, where $t$ is the elapsed time and $T_{1/2}$ is the half - life. Given $t = 1.022\times10^{10}$ years and $T_{1/2}=1.277\times10^{9}$ years.
$n=\frac{1.022\times 10^{10}}{1.277\times 10^{9}}=\frac{1.022}{1.277}\times10^{10 - 9}\approx8$

Step2: Use the radioactive decay formula

The radioactive decay formula is $N = N_0\times(\frac{1}{2})^n$, where $N_0$ is the initial amount and $N$ is the final amount. Here, $N_0 = 500.3$ g and $n = 8$.
$N=500.3\times(\frac{1}{2})^8=500.3\times\frac{1}{256}\approx1.95$ g

Answer:

Approximately 1.95 g