Solve the problem.
-In the formula $\mathrm { A } = \mathrm { Ie } \mathrm { kt } , \mathrm { A }$ is the amount of radioactive material remaining from an initial amount I 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 $0.175 \%$ annually. Determine the half-life of this isotope, to the nearest year.
A) $286 \mathrm { yr }$
B) $172 \mathrm { yr }$
C) $396 \mathrm { yr }$
D) $4 \mathrm { yr }$

Question

Quizplus · Accepted Answer

The decay rate is $0.175\%$, so $k = -0.00175$. The half-life is found by solving $0.5 = e^{-0.00175t}$. Solving for $t$ gives approximately 396 years.

Solve the Problem A=Iekt,A\mathrm { A } = \mathrm { Ie } \mathrm { kt } , \mathrm { A }A=Iekt,A

Solve the Problem $\mathrm { A } = \mathrm { Ie } \mathrm { kt } , \mathrm { A }$