Question

Solve the problem. In the formula A(t) = A0ekt, A(t) is the amount of radioactive material remaining from an initial amount A0 at a given time t and k is a negative constant determined by the nature of the material. A certain radioactive isotope has a half-life of approximately 1,950 years. How many years would be required for a given amount of this isotope to decay to 75% of that amount?

Answer 1

To determine the time it takes for a radioactive isotope to decay to 75% of its initial amount given a half-life of 1,950 years, we use the half-life formula A(t) = A0(1/2)^(t/T1/2) and solve for time t.

Step-by-step explanation:

The student posed a question regarding the time it would take for a given amount of a radioactive isotope with a half-life of 1,950 years to decay to 75% of its original amount. We'll use the half-life concept to solve this problem.

To find out the time required for the substance to reduce to 75%, we first recognize that after one half-life, we would have 50% of the substance remaining. To get to 75%, we need less than one half-life. The formula to calculate the time (t) is derived from the general equation for radioactive decay: A(t) = A0(1/2)^(t/T1/2), where A(t) is the amount at time t, A0 is the initial amount, T1/2 is the half-life, and t is the time in years.

Solving for t when A(t)/A0 = 0.75 (which is 75%), we have 0.75 = (1/2)^(t/1950). Taking the natural log of both sides, we get ln(0.75) = (t/1950)ln(1/2). Solving for t, we get t = 1950(ln(0.75)/ln(1/2)) years.

Therefore, we calculate the time it takes for the radioactive isotope to decay to 75% of its initial amount using the concept of half-lives and the natural logarithm.