Question

the function (m(h)=50(\frac{1}{2})^{\frac{h}{5.3}}) gives the mass, (m), of a radioactive substance remaining after (h) half - lives. cobalt - 60 has a half - life of about 5.3 years. which equation gives the mass of a 50 mg cobalt - 60 sample remaining after 10 years, and approximately how many milligrams remain? (m(1.887)=50(0.877)^{1.887}approx13.5) mg (m(1.887)=50(0.933)^{1.887}approx25) mg (m(0.5)=50(0.5)^{0.5}) (m(0.1887)=50(0.1887)^{0.1887}approx0) mg

Explanation:

Step1: Calculate number of half - lives

The half - life of Cobalt - 60 is 5.3 years. For 10 years, the number of half - lives $h=\frac{10}{5.3}\approx1.887$.

Step2: Use decay formula

The formula for radioactive decay is $A(x)=m(\frac{1}{2})^h$, where $m = 50$ mg (initial mass) and $h$ is the number of half - lives. Substituting $m = 50$ and $h\approx1.887$ into the formula, we get $A(x)=50(\frac{1}{2})^{1.887}$.
$(\frac{1}{2})^{1.887}=2^{- 1.887}\approx0.27$. Then $A(x)=50\times0.27 = 13.5$ mg.

Answer:

$A(x)=50(0.877)^x\approx13.5$ mg corresponds to the correct amount remaining after 10 years. So the answer is the option where $A(x)=50(0.877)^x$; $13.5$ mg.