Question

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.

Answer 1

So, what do we know about radioactive decays?

Answer 2

the graph is an exponential decay

Answer 3

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

Answer 4

is it -e^?

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Alright, have you had differential equations?

Answer 11

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

Answer 12

No

Answer 13

No
Not at all.

Answer 14

Yes i know, but why no, please?

Answer 15

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

Answer 16

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.

Answer 17

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

Answer 18

No.

Answer 19

So the rate itself is changing

Answer 20

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

Answer 21

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

Answer 22

but as vandreigan says, the rate is not constant.

Answer 23

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?

Answer 24

There are some things you'll need to know that aren't readily apparent, however. A half-life equals Tln2. Find T.

\[N(t) = N _{0}e ^{-t/T}\]

Answer 25

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.

Answer 26

ok. Thanks All!

Answer 27

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.

Answer 28

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

Answer 29

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.

Answer 30

sorry right*

Answer 31

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

Answer 32

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!

Answer 33

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

Answer 34

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.