Question

The half-life of iron-52 is approximately 8.3 hours. Step 1 of 3: Determine a so that A(t)=A0at describes the amount of iron-52 left after t hours, where A0 is the amount at time t=0. Round to six decimal places.

Answer 1

Answer:

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Step-by-step explanation:

Given the half like of a material to be 8.3 hours and the amount of iron-52 left after t hours is modeled by the equation $A(t) = A_0 a^{t}$, we can get A(t) as shown;

At t = 8.3 hours, A(8.3) = 1/2

Initially at t = 0; A(0) = 1

Substituting this values into the function we will have;

$\frac{1}{2} = 1 * a^{8.3}\\\\Taking \ the \ log \ of\ both \ sides;\\\\log(\frac{1}{2} ) = log(a^{8.3} )\\\\log(\frac{1}{2} ) = 8.3 log(a)\\\\\fr-0.30103 = 8.3 log(a)\\\Dividing\ both\ sides\ by \ 8.3\\\\\frac{-0.30103}{8.3} = log(a)\\\\log(a) = - 0.03627\\\\a =10^{-0.03627} \\\\a = 0.919878 (to\ 6dp)$