Question

The amount of a radioactive substance remaining after t years is given by the function f(t)=m(0.5)t/h, where m is the initial mass and h is the half-life in years. 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?

Answer 1

$f(10) = 50(0.5)^{\frac{10}{5.3}}$

f(10) = 13.52 mg

Step-by-step explanation:

The amount of a radioactive substance remaining after t years is given by the function $f(t) = m(0.5)^{\frac{t}{h}}$ ............ (1),

where m is the initial mass and h is the half-life in years.

Now, for Cobalt-60, h = 5.3 years, m = 50 mg and t = 10 years, then from equation (1) we get,

$f(10) = 50(0.5)^{\frac{10}{5.3}}$.

Hence, this is the required equation and from it, we get, f(10) = 13.52 mg (Answer)

Answer 2

Answer:

"f(10) = 50(0.5) 10/5.3; 13.5 mg" as of A for your correct answer.

Step-by-step explanation:

Here given Cobalt isotope has a half-life of 5.3 years.

The Initial amount is 50 mg.

Given time(10 years) is rough equals two half-lives.

So, after 5.3 years, 25 mg of isotope will left further in the next half-life only around 12.5 mg remains.

Among this option, 13.5 mg is the best possible.