anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
So, what do we know about radioactive decays?
anonymous
  • anonymous
the graph is an exponential decay
anonymous
  • anonymous
Indeed. Do you know the basic formula? Or how to derive it?

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anonymous
  • anonymous
is it -e^?
anonymous
  • anonymous
Well, it will end up being a function involving e with a negative exponent, but there's more to it.
anonymous
  • 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
  • anonymous
Every 3.3 hours, the amount of S is reduced by half.
anonymous
  • anonymous
no we were expected to derive it but its from last ur so i cant seem to recall it. haha
anonymous
  • anonymous
So how much of \(S_0\) is there after 3.3 hours?
anonymous
  • anonymous
Alright, have you had differential equations?
anonymous
  • anonymous
It should be \[3.3So/2t\]
anonymous
  • anonymous
No
anonymous
  • anonymous
No Not at all.
anonymous
  • anonymous
Yes i know, but why no, please?
anonymous
  • anonymous
Every 3.3 hours you are multiplying \(S_0\) by a factor of 1/2.
anonymous
  • 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
  • anonymous
So \[S = S_0 (\frac{1}{2})^{\frac{t}{3.3}}\]
anonymous
  • anonymous
No.
anonymous
  • anonymous
So the rate itself is changing
anonymous
  • anonymous
Indeed, the rate changes as the material decays. Which is reflected with an exponential decay.
anonymous
  • anonymous
Oops, polpak, that's what i wanted to write! sorry
anonymous
  • anonymous
but as vandreigan says, the rate is not constant.
anonymous
  • 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
  • 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
  • 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
  • anonymous
ok. Thanks All!
anonymous
  • 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
  • anonymous
What you wrote has no bearing on the question itself. I'm sorry, but its tangential to the proper approach.
anonymous
  • 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
  • anonymous
sorry right*
anonymous
  • anonymous
It's an average rate. Decays are rooted in probability.
anonymous
  • 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
  • anonymous
Good night, its 1:00 AM here. toodles!
anonymous
  • 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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