Question

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.

Answer 1

693 Years

Explanation:

Given an initial amount
`A_o` 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:

`A(t)= A_oe^(kt)`

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

`(1)/(2) A_o= A_oe^(-0.001t)\\e^(-0.001t)=(1)/(2)\\$Take the natural logarithm of both sides\\-0.001t=ln(0.5)\\t=ln(0.5) / -0.001\\t=693.15\approx 693 \:years`

The half-life of the isotope is 693 years.