A certain i radioactive substance has a known half-life of 18.4 years. Scientist begins with 100 g of the substance. Find what happens in 10 years.

Question

Please, can you use A=A0e^kt. Thank you

Answer 1

A = a₀(0.5)^t/h

A = 100(0,5)^10/18.4

A = 100(0,5)^0.543

A = 100(0.68634)

A = 68.634

Answer 2

A = A₀e^kt;

^{If t = 18.4 years then A = 1/2A}0^{. We get exponential equation for k: 1/2 = e^(18.4·k),}

^{from here k = ln(1/2)/18.4 = - .03767 and our decay law is A = 100e^(-.03767t).}

^{In ten years (t = 10), A 100e^(-..03767·10) = 68.61 g.}