SOLUTION: The amount of radioactive tracer remaining after t days is given by A =Ao e-o.o58t, where AO is the starting amount at the beginning of the time period. How many days will it take
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-> SOLUTION: The amount of radioactive tracer remaining after t days is given by A =Ao e-o.o58t, where AO is the starting amount at the beginning of the time period. How many days will it take
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Question 72953: The amount of radioactive tracer remaining after t days is given by A =Ao e-o.o58t, where AO is the starting amount at the beginning of the time period. How many days will it take for one half of the original amount to decay?
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Thanks Found 2 solutions by bucky, jim_thompson5910:Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! Given:
.
.
A is the amount of material at time t. is the starting amount of material. And
t is the amount of elapsed time in days. You are asked to find the amount of time t that it
will take for the amount of material to equal half the amount you started with.
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Since you started with when it is half gone the A will equal
.
Substituting this in for A results in the equation becoming:
.
.
If you divide both sides of this equation by the term on both sides
drops out and you are left with:
.
.
Take the logarithm of both sides. Since the equation has an e in it, take the natural logarithm,
ln:
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ln(1/2) = ln(e^(-0.058*t))
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On a calculator you will find that ln(1/2) = ln(0.5) = -0.69314718. Substitute this into
the equation for ln(1/2) to get:
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-0.69314718 = ln(e^(-0.058*t))
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On the right side, by the rules of logarithms in a logarithm of a term with an exponent
the exponent can be taken as the multiplier of the logarithm. As a result, the term:
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ln(e^(-0.058*t))
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is equal to the term:
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(-0.058*t)*ln(e)
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but ln(e) is equal to 1. So this term further reduces to (-0.058*t). Substitute this
into the equation and you now have:
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-0.69314718 = (-0.058*t)
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Multiply both sides by -1 to eliminate the negative signs and have:
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0.69314718 = 0.058*t
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Finally divide both sides by 0.058 to solve for t:
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0.69314718/0.058 = t
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And after the division is performed the equation becomes:
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11.9508 = t
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So after about 12 days one-half of the original material has decayed.
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Hope this helps you to decipher the problem.
You can put this solution on YOUR website! If is the original amount, then is half of the original amount. So set equal to Now solve for t (remember e is a constant number) Divide both sides by Now take the natural log of both sides (this undoes the natural base e) Divide both sides by -0.058
Since -ln(1/2)=0.693147 approximately
So after 11.95 days or so the original amount remaining is half. Hope that helps.
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