the half life of lead is 3.3 hours. The initial amount of the substance if S (subscript) 0 grams. Express the amount of substance S remaining as a function of time t.

- anonymous

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- anonymous

So, what do we know about radioactive decays?

- anonymous

the graph is an exponential decay

- anonymous

Indeed. Do you know the basic formula? Or how to derive it?

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## More answers

- anonymous

is it -e^?

- anonymous

Well, it will end up being a function involving e with a negative exponent, but there's more to it.

- anonymous

I'm not certain what level of math you are at, so I'm just wondering if you were given a formula, or if you are expected to derive it.

- anonymous

Every 3.3 hours, the amount of S is reduced by half.

- anonymous

no we were expected to derive it but its from last ur so i cant seem to recall it. haha

- anonymous

So how much of \(S_0\) is there after 3.3 hours?

- anonymous

Alright, have you had differential equations?

- anonymous

It should be \[3.3So/2t\]

- anonymous

No

- anonymous

No
Not at all.

- anonymous

Yes i know, but why no, please?

- anonymous

Every 3.3 hours you are multiplying \(S_0\) by a factor of 1/2.

- anonymous

Radioactive decay is a random time process where, on average, half of the substance will decay in the half-life time. It's an exponential decay that will reference the amount present, the half-life time, and the time elapsed.

- anonymous

So
\[S = S_0 (\frac{1}{2})^{\frac{t}{3.3}}\]

- anonymous

No.

- anonymous

So the rate itself is changing

- anonymous

Indeed, the rate changes as the material decays. Which is reflected with an exponential decay.

- anonymous

Oops, polpak, that's what i wanted to write! sorry

- anonymous

but as vandreigan says, the rate is not constant.

- anonymous

So Vandreigan: Isn't what polpak wrote is the rate at every instant?
Integrating it would do the job, isnt it? Or i'm missing something?

- anonymous

There are some things you'll need to know that aren't readily apparent, however. A half-life equals Tln2. Find T.
\[N(t) = N _{0}e ^{-t/T}\]

- anonymous

So, since you are using S as the mass, just change N to S, find T, plug it all in. If you need to derive this equation, let me know.

- anonymous

ok. Thanks All!

- anonymous

What I wrote was correct.. If you want to rewrite \[(\frac{1}{2})^{\frac{1}{3.3}}\]
as a power of e that's fine, but either way it'll be the same number.

- anonymous

What you wrote has no bearing on the question itself. I'm sorry, but its tangential to the proper approach.

- anonymous

No, according to you, in 3.3 hours the material should half decay, but its not the case.
What i think its the rate at every instant, I'm not sure if its write or wrong.

- anonymous

sorry right*

- anonymous

It's an average rate. Decays are rooted in probability.

- anonymous

Anyway, have a good afternoon. There is plenty of information online involving the solutions to the decay ODE's and the like. Most physics books will cover it pretty well as well!

- anonymous

Good night, its 1:00 AM here. toodles!

- anonymous

So \(t_{1/2} = \tau \cdot ln2 \)\[\implies 3.3 = \tau \cdot ln2 \]\[\implies \frac{1}{\tau} = \frac{ln2}{3.3} \]\[ \implies \frac{-1}{\tau} = \frac{-ln2}{3.3}\]\[\implies e^{\frac{-1}{\tau}} = (e^{ln2})^{\frac{-1}{3.3}} = 2^{\frac{-1}{3.3}} = (\frac{1}{2})^{\frac{1}{3.3}}\]
Which is exactly what I said. I understand the relationship between exponential decay and ODE's, but this student is coming from an algebraic background and won't really benifit from your explanation as much as just setting up the problem in a straightforward way.

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