Question 126105: The formula for describing the amount of radioactive Plutonium-35 is A=Ao(1/2)(^t/4370),where A is the remaining amount of Plutonium-35 and t is the time given in years.
a) When will the remaining Plutonium-35 be 1/8 of the original amount?
b) When will the remaining Plutonium-35 be 1/sqrt2 of the original amount?
c) When will the remaining Plutonium-35 be 1/sqrt32 of the original amount?
I am not sure if I have typed the Formala A=Ao(1/2)^t/4370 correctly. I don't know if I should say A= Abase0(1/2)(^t/4370).Can someone clarify this for me?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The formula for describing the amount of radioactive Plutonium-35 is
A(t) = Ao(1/2)^(t/4370),where A is the remaining amount of Plutonium-35 and t is the time given in years
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This formula means:
A(t) is the amount of the plutonium at time "t"
Ao is the amount of plutonium you start with
(1/2) is the decrease rate
t/4370 means the half-life is 4370 yearsevery 4370 years you multiply by (1/2)
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I'll do part "a" and leave the others to you.
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a) When will the remaining Plutonium-35 be 1/8 of the original amount?
(1/8)Ao = Ao(1/2)^(t/4370)
(1/8) = (1/2)^(t/4370)
Take the log of both sides to get:
log(1/8) = (t/4370)log (1/2)
t/4370 = [log(1/8)/log(1/2)] = 3
t = 13110 yrs
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Cheers,
Stan H.
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b) When will the remaining Plutonium-35 be 1/sqrt2 of the original amount?
c) When will the remaining Plutonium-35 be 1/sqrt32 of the original amount?
I am not sure if I have typed the Formala A=Ao(1/2)^t/4370 correctly. I don't know if I should say A= Abase0(1/2)(^t/4370).Can someone clarify this for me?
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