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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.
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, what do we know about radioactive decays?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the graph is an exponential decay

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Indeed. Do you know the basic formula? Or how to derive it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, it will end up being a function involving e with a negative exponent, but there's more to it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0Every 3.3 hours, the amount of S is reduced by half.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no we were expected to derive it but its from last ur so i cant seem to recall it. haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So how much of \(S_0\) is there after 3.3 hours?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, have you had differential equations?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It should be \[3.3So/2t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes i know, but why no, please?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Every 3.3 hours you are multiplying \(S_0\) by a factor of 1/2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Radioactive decay is a random time process where, on average, half of the substance will decay in the halflife time. It's an exponential decay that will reference the amount present, the halflife time, and the time elapsed.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So \[S = S_0 (\frac{1}{2})^{\frac{t}{3.3}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the rate itself is changing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Indeed, the rate changes as the material decays. Which is reflected with an exponential decay.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oops, polpak, that's what i wanted to write! sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but as vandreigan says, the rate is not constant.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0There are some things you'll need to know that aren't readily apparent, however. A halflife equals Tln2. Find T. \[N(t) = N _{0}e ^{t/T}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, 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
 5 years ago
Best ResponseYou've already chosen the best response.0What 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
 5 years ago
Best ResponseYou've already chosen the best response.0What you wrote has no bearing on the question itself. I'm sorry, but its tangential to the proper approach.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, 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
 5 years ago
Best ResponseYou've already chosen the best response.0It's an average rate. Decays are rooted in probability.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Anyway, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Good night, its 1:00 AM here. toodles!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So \(t_{1/2} = \tau \cdot ln2 \)\[\implies 3.3 = \tau \cdot ln2 \]\[\implies \frac{1}{\tau} = \frac{ln2}{3.3} \]\[ \implies \frac{1}{\tau} = \frac{ln2}{3.3}\]\[\implies e^{\frac{1}{\tau}} = (e^{ln2})^{\frac{1}{3.3}} = 2^{\frac{1}{3.3}} = (\frac{1}{2})^{\frac{1}{3.3}}\] 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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