the amount of a sample remaining after t days is given by the equation $p(t)=a(\frac{1}{2})^{\frac{t}{h}}$…

the amount of a sample remaining after t days is given by the equation $p(t)=a(\frac{1}{2})^{\frac{t}{h}}$, where a is the initial amount of the sample and h is the half - life, in days, of the substance. a sample contains 18% of its original amount of radon - 222. the half - life of radon - 222 is about 3.8 days. which is the best estimate for the age of the sample?\no 1.5 days\no 2.5 days\no 9.4 days\no 21.1 days

the amount of a sample remaining after t days is given by the equation $p(t)=a(\frac{1}{2})^{\frac{t}{h}}$, where a is the initial amount of the sample and h is the half - life, in days, of the substance. a sample contains 18% of its original amount of radon - 222. the half - life of radon - 222 is about 3.8 days. which is the best estimate for the age of the sample?\no 1.5 days\no 2.5 days\no 9.4 days\no 21.1 days

Answer

Explanation:

Step1: Set up the equation

We know that $P(t)=0.18A$ (since the sample contains 18% of its original amount), $h = 3.8$ days, and the formula $P(t)=A\left(\frac{1}{2}\right)^{\frac{t}{h}}$. Substituting the values, we get $0.18A=A\left(\frac{1}{2}\right)^{\frac{t}{3.8}}$. Since $A\neq0$, we can divide both sides by $A$ to obtain $0.18=\left(\frac{1}{2}\right)^{\frac{t}{3.8}}$.

Step2: Take the natural - logarithm of both sides

$\ln(0.18)=\ln\left(\left(\frac{1}{2}\right)^{\frac{t}{3.8}}\right)$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we have $\ln(0.18)=\frac{t}{3.8}\ln\left(\frac{1}{2}\right)$.

Step3: Solve for $t$

First, note that $\ln(0.18)\approx - 1.7148$ and $\ln\left(\frac{1}{2}\right)\approx-0.6931$. Then, $t = 3.8\times\frac{\ln(0.18)}{\ln\left(\frac{1}{2}\right)}$. Substitute the values of the logarithms: $t = 3.8\times\frac{- 1.7148}{-0.6931}\approx3.8\times2.4741\approx9.4$ days.

Answer:

9.4 days