Question

The function f(h)=m(1/2)^h gives the mass, m, of a radioactive substance remaining after h half-lives. Cobalt-60 has a half-life

Answer 1

Answer:

The answer is C

Step-by-step explanation:

I took the test

Answer 2

Answer:

Third choice:

                $f(x)=50(0.877)^{10};13.5mg$

Explanation:

1. Data:

• Function:

                $f(h)=m(1/2)^h$

• Cobalt-60's half-life: 5.3 years

• Initial mass of cobalt-60:  50 mg

2. Unknown:

• equation for the mass of cobalt-60 remaining after 10 years = ?

• mass remaining =?

3. Solution

Half-life is the time it takes a sample to  decay to half of its initial amount. It is considered constant. Hence, when one half-life passes, the sample has decayed to 50% of the original amount; when two half-lives pass, the sample has decayed to (1/2)×(1/2) = 1/4 = 25%; when three half-lives have elapsed, the sample has decayed to (1/2)³ = 1/8 = 12.5% of its original amount, and so on.

Then, the amound of a sample remaining is calculated as the original amount times (1/2) raised to the number of half-lives elapsed, which is what the given function,  $f(h)=m(1/2)^h$ models.

You just must substitute the data into the function to get the answer to the question:

            $f(x)=50(0.5)^{(10/5.3)}$

Where, 50 is the original mass of 50g, 0.5 is equal to 1/2, and 10/5.3 gives the number of half-lives (the number of times that 5.3 years is contained in 10 years).

Simplifying:

$f(x)=50(0.5)^{(10/5.3)}=50((0.5)^{(1/5.3})^{10}\approx50(0.877)^{10}$

Which corresponds to the third choice of the list.

Computing:

               $f(x)\approx50(0.877)^{10}=13.5$

Which also corresponds to the third choice.