Select the correct answer.The amount of a radioactive substance remaining as it decays over time is 𝐴=𝐴0⁢(0.5)𝑡ℎ , where A represents the final amount, 𝐴0 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. 19,035 years C. 28,650 years D. 8,267 years

To determine how many years it will take for carbon-14 to decay to 10 percent of its original amount, we can use the given formula:

$A = A_0 \cdot (0.5)^{\frac{t}{h}}$

where:

• $A$ is the final amount,
• $A_0$ is the original amount,
• $t$ is the number of years,
• $h$ is the half-life of the substance.

Given:

• The half-life $h$ of carbon-14 is 5,730 years.
• We want to find $t$ when $A = 0.1 \cdot A_0$.

Let's substitute these values into the formula and solve for $t$:

$0.1 \cdot A_0 = A_0 \cdot (0.5)^{\frac{t}{5730}}$

First, divide both sides by $A_0$:

$0.1 = (0.5)^{\frac{t}{5730}}$

Next, take the natural logarithm (ln) of both sides to solve for $t$:

$\ln(0.1) = \ln\left((0.5)^{\frac{t}{5730}}\right)$

Using the property of logarithms $\ln(a^b) = b \cdot \ln(a)$:

$\ln(0.1) = \frac{t}{5730} \cdot \ln(0.5)$

Now, solve for $t$:

$t = \frac{\ln(0.1)}{\ln(0.5)} \cdot 5730$

Calculate the values of the logarithms:

$\ln(0.1) \approx -2.3026$ $\ln(0.5) \approx -0.6931$

Substitute these values back into the equation:

$t = \frac{-2.3026}{-0.6931} \cdot 5730$

$t \approx 3.3219 \cdot 5730$

$t \approx 19,035 \text{ years}$

Therefore, the correct answer is:

B. 19,035 years