exponential decay

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May 20th 2008, 08:37 PM
keemariee

exponential decay

Five grams of a certain radioactive isotope decay to three grams in 100 years. Assuming that the rate of decay is proportional to the amount present, after how many more years will there be just one gram?

Now assume that the rate of decay is proportional to the square of the amount present. After how many more years will there be just one gram?

i found 315 and 600 but theyre wrong. any help?
May 20th 2008, 08:46 PM
Mathstud28

Quote:

Originally Posted by keemariee

Five grams of a certain radioactive isotope decay to three grams in 100 years. Assuming that the rate of decay is proportional to the amount present, after how many more years will there be just one gram?

Now assume that the rate of decay is proportional to the square of the amount present. After how many more years will there be just one gram?

i found 315 and 600 but theyre wrong. any help?

We know that since it is proportional to itself it will follow the formula

$y=Ce^{kt}$

We see that t=0 y=5

So $5=Ce^{5\cdot{0}}\Rightarrow{C=5}$

Next we know that at t=100 y=3

So $3=5e^{100k}$

so $\ln\bigg(\frac{3}{5}\bigg)=100k\Rightarrow{k=\ln\b igg(\sqrt[100]{\frac{3}{5}}\bigg)}$

So we know have

$y=5e^{\ln\bigg(\sqrt[100]{\frac{3}{5}}\bigg)t}$

and you want to solve $1=5e^{\ln\bigg(\sqrt[100]{\frac{3}{5}}\bigg)t}$

how would you do that?
May 21st 2008, 04:23 AM
topsquark

Quote:

Originally Posted by keemariee

Five grams of a certain radioactive isotope decay to three grams in 100 years. Assuming that the rate of decay is proportional to the amount present, after how many more years will there be just one gram?

Now assume that the rate of decay is proportional to the square of the amount present. After how many more years will there be just one gram?

i found 315 and 600 but theyre wrong. any help?

You mostly have the answer for the first one correct. Re-read the question: "...after how many more years will there be just one gram?" So your answer would be 315 - 100 = 215 years. This is the same error that you made in the second problem.

-Dan