Question

select the correct answer. the amount of a radioactive substance remaining as it decays over time is $a = a_0(0.5)^{\frac{t}{h}}$, where $a$ represents the final amount, $a_0$ represents the original amount, $t$ represents the number of years, and $h$ represents the half - life of the substance. carbon - 14 is a radioactive isotope that has a half - life of 5,730 years. approximately how many years will it take for carbon - 14 to decay to 10 percent of its original amount?
a. 16,396 years
b. 28,650 years
c. 8,267 years
d. 19,035 years

Explanation:

Step1: Set up the equation

We know that $A = 0.1A_0$ (since the substance decays to 10% of its original amount) and $h = 5730$. Substitute into the decay - formula $A=A_0(0.5)^{\frac{t}{h}}$. So, $0.1A_0=A_0(0.5)^{\frac{t}{5730}}$. Divide both sides by $A_0$ (since $A_0
eq0$), we get $0.1=(0.5)^{\frac{t}{5730}}$.

Step2: Take the natural logarithm of both sides

$\ln(0.1)=\ln((0.5)^{\frac{t}{5730}})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we have $\ln(0.1)=\frac{t}{5730}\ln(0.5)$.

Step3: Solve for $t$

First, we know that $\ln(0.1)\approx - 2.3026$ and $\ln(0.5)\approx-0.6931$. Then, $t = 5730\times\frac{\ln(0.1)}{\ln(0.5)}$. Substitute the values of the logarithms: $t = 5730\times\frac{-2.3026}{-0.6931}$. Calculate $\frac{-2.3026}{-0.6931}\approx3.322$. Then $t = 5730\times3.322\approx19035$.

Answer:

D. 19,035 years