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

Answer:

693 Years

Step-by-step 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.

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

The half-life of the isotope is 693 years.