Question

the function (m(h)=m_0(\frac{1}{2})^h) gives the mass, m, of a radioactive substance remaining after h half - lives. iron has a half - life of 2.7 years. which equation gives the mass of a 200 mg iron sample remaining after y years, and approximately how many milligrams remain after 12 years?

(f(x)=2.7(0.5)^{200y}, 1.7 mg)
(f(x)=200(0.5)^{12})
(f(x)=200(0.185)^{12}, 3.2 mg)
(f(x)=200(0.774)^{12}, 9.2 mg)

Explanation:

Step1: Find the number of half - lives

The half - life of iron is $t_{1/2}=2.7$ years. The number of half - lives $h$ in $y$ years is $h = \frac{y}{2.7}$. The initial mass $m_0 = 200$ mg. The formula for the remaining mass $m(y)$ of a radioactive substance is $m(y)=m_0(\frac{1}{2})^{\frac{y}{2.7}}$. Substituting $m_0 = 200$, we get $m(y)=200\times(0.5)^{\frac{y}{2.7}}$.

Step2: Calculate the remaining mass after 12 years

When $y = 12$, we have $h=\frac{12}{2.7}\approx4.44$. Then $m(12)=200\times(0.5)^{4.44}$.
$(0.5)^{4.44}=0.5^{4}\times0.5^{0.44}$. We know that $0.5^{4}=\frac{1}{16}=0.0625$. Using a calculator, $0.5^{0.44}\approx0.74$. So $(0.5)^{4.44}\approx0.0625\times0.74 = 0.04625$. Then $m(12)=200\times0.04625 = 9.25\approx9.2$ mg. The correct equation is $m(y)=200\times(0.74)^{\frac{y}{2.7/ \log_{0.74}0.5}}=200\times(0.74)^{y}$ (since $a = b^{\log_{b}a}$ and $\log_{0.74}0.5=\frac{\log0.5}{\log0.74}\approx2.7$).

Answer:

$m(x)=200(0.74)^{x};9.2$ mg