Question

The amount of radioactive substance remaining after two years is given by the function f(t) = (0.5)^t/h , where h is the half-life. If ⁶⁰Co has a half-life of 5.3 years, which equation represents the mass of 50 mg ⁶⁰Co after 10 years, and approximately how many milligrams remain?

a) f(10) = 50 × (0.5)^10/5.3 , ~37.74 mg remain
b) f(10) = 50 × (0.5)^5.3/10 , ~27.43 mg remain
c) f(10) = 50 × (0.5)^10/5.3 , ~27.43 mg remain
d) f(10) = 50 × (0.5)^5.3/10 , ~37.74 mg remain

Answer 1

The correct equation to represent the mass of 50 mg of ⁶⁰Co after 10 years is f(10) = 50 × (0.5)^10/5.3, and approximately 17.435 mg remains.

Step-by-step explanation:

To find the equation that represents the mass of 50 mg of ⁶⁰Co after 10 years using its half-life, we plug in the half-life (5.3 years) and the elapsed time (10 years) into the given decay function, f(t) = (0.5)^t/h. The correct expression for the mass remaining after 10 years is f(10) = 50 × (0.5)^¹⁰/⁵.₃, which corresponds to option (c).

Now, we calculate the actual mass remaining:

f(10) = 50 × (0.5)^{¹⁰/⁵.₃} ≈ 50 × (0.5)^1.887 ≈ 50 × 0.3487 ≈ 17.435 mg

Therefore, approximately 17.435 mg of ⁶⁰Co remains after 10 years.

The equation that represents the mass of 50 mg ⁶⁰Co after 10 years is f(10) = 50 × (0.5)^10/5.3. To calculate the approximate amount of milligrams that remain, we can substitute the value of t into the equation. Evaluating the expression, we find that approximately 37.74 mg of ⁶⁰Co remain.