SOLUTION: STANBON You helped me with this problem:
The level of thorium in a sample decreases by a factor of one-half every 2 million years. A meteorite is discovered to have only 8.6% of
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Question 195268This question is from textbook
: STANBON You helped me with this problem:
The level of thorium in a sample decreases by a factor of one-half every 2 million years. A meteorite is discovered to have only 8.6% of its original thorium remaining. How old is the meteorite? I already have the answer, but need to know
What does Ao represent? Do you know what the value of Ao? What does t represent? What are the units of t? Thanks!
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A(t) = Ao*(1/2)^(t/2*10^6)
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0.086Ao = Ao*(1/2)^(t/(2 million))
(1/2)^(t/2 million) = 0.086
Take the log to get:
(t/2 million) = log(0.086)/log(1/2) = 3.53951953
t = (2 million)*(3.53951953)
t = 7,079,039.06
This question is from textbook
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
You helped me with this problem:
The level of thorium in a sample decreases by a factor of one-half every 2 million years. A meteorite is discovered to have only 8.6% of its original thorium remaining. How old is the meteorite? I already have the answer, but need to know
What does Ao represent? Do you know what the value of Ao? What does t represent? What are the units of t? Thanks!
-----------
A(t) = Ao*(1/2)^(t/2*10^6)
A(t) is the amount of material you have at time "t".
Ao is the amount of material you have in the beginning: at time t=0
In your problem you do not know the amount at time t=0, so you
do not know Ao. But since you have it on both sides of the equation,
it cancels out.
In your problem t represets the number of years the thorium has
been around (presumably in the meteorite).
-----------------------
0.086Ao = Ao*(1/2)^(t/(2 million))
(1/2)^(t/2 million) = 0.086
Take the log to get:
(t/2 million) = log(0.086)/log(1/2) = 3.53951953
t = (2 million)*(3.53951953)
t = 7,079,039.06 years
==============================
Cheers,
Stan H.
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