In the formula A(t) $\mathrm { A } ( \mathrm { t } ) = \mathrm { A } _ { 0 } \mathrm { e } ^ { k t }$ , A is the amount of radioactive material remaining from an initial amount $\mathrm{A}_{0}$ 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 0.125% annually. Determine the half-life of this isotope, to the nearest year.
A) 400 years
B) 241 years
C) 6 years
D) 555 years

Question

Quizplus · Accepted Answer

The decay rate is 0.125% annually, which means k = -0.00125. The half-life is found using the formula $ t_{1/2} = \frac{\ln(2)}{-k} $. Calculating gives $ t_{1/2} \approx \frac{0.693}{0.00125} \approx 554.4 $, which rounds to 555 years.

In the Formula A(t) A(t)=A0ekt\mathrm { A } ( \mathrm { t } ) = \mathrm { A } _ { 0 } \mathrm { e } ^ { k t }A(t)=A0​ekt

In the Formula A(t) $\mathrm { A } ( \mathrm { t } ) = \mathrm { A } _ { 0 } \mathrm { e } ^ { k t }$