In the formula A(t)equals=Upper A0A0 Superscript ktkt, A is the amount of radioactive material remaining from an initial amount Upper A 0A0 at a given time t, and k is a negative constant determined by the nature of the material. A certain radioactive isotope decays at a rate of 0.10.1% annually. Determine the half-life of this isotope, to the nearest year.

Given an initial amount [tex]A_o[/tex] and k (a negative constant) determined by the nature of the material, the amount of radioactive material remaining at a given time t, is determined using he formula:

[tex]A(t)= A_oe^{kt}[/tex]

If a certain radioactive isotope decays at a rate of 0.1% annually.

[tex]\frac{1}{2} A_o= A_oe^{-0.001t}\\e^{-0.001t}=\frac{1}{2}\\$Take the natural logarithm of both sides\\-0.001t=ln(0.5)\\t=ln(0.5) \div -0.001\\t=693.15\approx 693 \:years[/tex]

The half-life of the isotope is 693 years.