SOLUTION: The half-life of gold-195m is approximately 30.5 seconds. Step 1 of 3 : Determine a so that A(t)=A0at describes the amount of gold-195m left after t seconds, where A0 is the am

Question 1140477: The half-life of gold-195m is approximately 30.5 seconds.
Step 1 of 3 : Determine a so that A(t)=A0at describes the amount of gold-195m left after t seconds, where A0 is the amount at time t=0. Round to six decimal places.

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
N=Noe^(-kt)
half-life is where N=1/2 No so N/No=1/2
ln (1/2)=-kt, since ln e^-kt is -kt
-0.693=-kt, but don't round yet.
t=30.5 in this instance
so 0.693=30.5 k
k=0.022726
A(t)=Aoe^-0.022726t

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