Question

A certain radioactive isotope has a half-life of 5 days. If one is to make a table showing the half-life decay of a sample of this isotope from 32 grams to 1 gram; list the time (in days, starting with t = 0) in the first column and the mass remaining (in grams) in the second column, which type of sequence is used in the first column and which type of sequence is used in the second column?

Answer 1

Do you know the formula for this sort of thing?

Answer 2

i know it has to do with geometric sequences, but idk how to apply it here

Answer 3

Well, in terms of a geo-series, I'm not sure the best way to write it, but we know that the original amount constantly gets cut in half. So starting from 32, we multiply that by (1/2) for each half-life gone by. So I would personally put:
\[32\sum_{n=1}^{\infty}(\frac{ 1 }{ 2 })^{n}\] Where n is the amount of half-lifes that have occured.

Answer 4

Apart from a geo-series, there is an exponential growth and decay formula that goes like this:
\[y = Ce ^{kt}\] C and k are constants that must be found and t is time. So for you, at t = 0 we would write the equation like this:
\[32 = Ce ^{k(0)}\] When we do that, we find out that C = 32. (C always equals the original amount). Now using that value we would need to find k. There's a quick formula for k, which is \[\frac{ -\ln(2) }{ half-life }\]So we have a value for C and a value for k, now we just need to see how much time has gone by before we finally get to the point we only have 1 gram left. Now it can be found easily logically, but if you had an awkward number we would need to do this:

\[1 = 32e ^{\frac{ -t(\ln2) }{ 5 }}\] That is just time t multplied by the formula for k I have above. So I would first divide both sides by 32 and get:
\[\frac{ 1 }{ 32 } = e ^{\frac{ -t(\ln2) }{ 5 }}\]Now I take the natural log of both sides and get:
\[\ln(\frac{ 1 }{ 32 }) = \frac{ -t(\ln2) }{ 5 }\]Multiply both sides by -(5/ln2) and then calculator work tells me t = 25 days ^_^

Answer 5

Thank you