so, i'm stuck on this question...please help me, thanks!
a radioactive material decays exponentially. the percent, P, of the material left after 't' years is given by P(t)=100(1.2)^-t.
a) determine the half-life of the substance
b)how fast is the substance decaying at the point where the half-life is reached?
a radioactive material decays exponentially. the percent, P, of the material left after 't' years is given by P(t)=100(1.2)^-t.
a) determine the half-life of the substance
b)how fast is the substance decaying at the point where the half-life is reached?
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a) The half life is the time required for the original amount (100) to decay to half that value (50).
50 = 100[1.2^(-t)]
0.5 = 1.2^(-t)
ln(0.5) = (-t)ln(1.2)
t = -ln(0.5)/ln(1.2)
= 0.693147181/0.182321557 = 3.801784017 years
b) Find the derivative
dP/dt = 100[1.2^(-t)]ln(1.2)(-1)
= -100ln(1.2)[1.2^(-t)]
When t = 3.801784017 years,
dP/dt = -100ln(1.2)[1.2^(-3.801784017)]
= -100(0.182321557)(0.5)
= -9.11607784 percent per year
50 = 100[1.2^(-t)]
0.5 = 1.2^(-t)
ln(0.5) = (-t)ln(1.2)
t = -ln(0.5)/ln(1.2)
= 0.693147181/0.182321557 = 3.801784017 years
b) Find the derivative
dP/dt = 100[1.2^(-t)]ln(1.2)(-1)
= -100ln(1.2)[1.2^(-t)]
When t = 3.801784017 years,
dP/dt = -100ln(1.2)[1.2^(-3.801784017)]
= -100(0.182321557)(0.5)
= -9.11607784 percent per year
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I'm stuck, too. Just for a moment, pretend the "t = 2" Than the percentage of the radioactive material left after two years = 100(1.2)^1/2. Okay. That would be something in the .7 - .9(100) area. To find the half-life you need to solve for "t" so that (1.2)^-t = .5 That number will give you answer of .50 or 50%. I have no idea how to find it.
For part b).... IMHO that is a carelessly worded question: You could say it is decaying half as fast as it was initially - because there are only half as many of the original radioactive atoms remaining. You could say it is decaying at the same rate as it was initially - because the half-life of that remaining radioactive material remains the same.
I'd go with the first suggested answer. You can give ground to the argument that there are only half as many neutrons being released, but stand firmly in your argument that the RATE of decay is the same.
For part b).... IMHO that is a carelessly worded question: You could say it is decaying half as fast as it was initially - because there are only half as many of the original radioactive atoms remaining. You could say it is decaying at the same rate as it was initially - because the half-life of that remaining radioactive material remains the same.
I'd go with the first suggested answer. You can give ground to the argument that there are only half as many neutrons being released, but stand firmly in your argument that the RATE of decay is the same.
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let half life = t years so after t years the substance left would be half i.e 50 percent
so, 50= 100(1.2)^-t
now you can calculate half life by using logarithms easily or calculator. w/e suitable to you.
to calculate the substance decaying at the point where half life is reached is calculate derivative of p with respect to t.
dp/dt = -100(1.2)^-t(log1.2)
now put value of t as obtained above in the expression and you get the rate of decay =)
so, 50= 100(1.2)^-t
now you can calculate half life by using logarithms easily or calculator. w/e suitable to you.
to calculate the substance decaying at the point where half life is reached is calculate derivative of p with respect to t.
dp/dt = -100(1.2)^-t(log1.2)
now put value of t as obtained above in the expression and you get the rate of decay =)
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a)
50 = 100*(1.2)^(-t)
log(50/100) = log(1.2^(-t))
log(50/100) = -t(log1.2)
log(50/100) / log(1.2) = -t
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t = -log(1/2) / log(1.2) years. If you have to come up with an approximation then t ≅ -3.801784017 years.
a)
50 = 100*(1.2)^(-t)
log(50/100) = log(1.2^(-t))
log(50/100) = -t(log1.2)
log(50/100) / log(1.2) = -t
.
t = -log(1/2) / log(1.2) years. If you have to come up with an approximation then t ≅ -3.801784017 years.