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anonymous

  • 5 years ago

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.

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  1. anonymous
    • 5 years ago
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    So, what do we know about radioactive decays?

  2. anonymous
    • 5 years ago
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    the graph is an exponential decay

  3. anonymous
    • 5 years ago
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    Indeed. Do you know the basic formula? Or how to derive it?

  4. anonymous
    • 5 years ago
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    is it -e^?

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

  6. anonymous
    • 5 years ago
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    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.

  7. anonymous
    • 5 years ago
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    Every 3.3 hours, the amount of S is reduced by half.

  8. anonymous
    • 5 years ago
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    no we were expected to derive it but its from last ur so i cant seem to recall it. haha

  9. anonymous
    • 5 years ago
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    So how much of \(S_0\) is there after 3.3 hours?

  10. anonymous
    • 5 years ago
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    Alright, have you had differential equations?

  11. anonymous
    • 5 years ago
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    It should be \[3.3So/2t\]

  12. anonymous
    • 5 years ago
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    No

  13. anonymous
    • 5 years ago
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    No Not at all.

  14. anonymous
    • 5 years ago
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    Yes i know, but why no, please?

  15. anonymous
    • 5 years ago
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    Every 3.3 hours you are multiplying \(S_0\) by a factor of 1/2.

  16. anonymous
    • 5 years ago
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    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.

  17. anonymous
    • 5 years ago
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    So \[S = S_0 (\frac{1}{2})^{\frac{t}{3.3}}\]

  18. anonymous
    • 5 years ago
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    No.

  19. anonymous
    • 5 years ago
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    So the rate itself is changing

  20. anonymous
    • 5 years ago
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    Indeed, the rate changes as the material decays. Which is reflected with an exponential decay.

  21. anonymous
    • 5 years ago
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    Oops, polpak, that's what i wanted to write! sorry

  22. anonymous
    • 5 years ago
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    but as vandreigan says, the rate is not constant.

  23. anonymous
    • 5 years ago
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    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?

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

  25. anonymous
    • 5 years ago
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    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.

  26. anonymous
    • 5 years ago
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    ok. Thanks All!

  27. anonymous
    • 5 years ago
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    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.

  28. anonymous
    • 5 years ago
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    What you wrote has no bearing on the question itself. I'm sorry, but its tangential to the proper approach.

  29. anonymous
    • 5 years ago
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    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.

  30. anonymous
    • 5 years ago
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    sorry right*

  31. anonymous
    • 5 years ago
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    It's an average rate. Decays are rooted in probability.

  32. anonymous
    • 5 years ago
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    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!

  33. anonymous
    • 5 years ago
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    Good night, its 1:00 AM here. toodles!

  34. anonymous
    • 5 years ago
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    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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