Question

The rate of decay of a radioactive substance is proportional to the amount of such substance present.
(a) Today we have Ao g of a radioactive substance. Given that one-third of the substance decay every 5 years, how much will be left t years from today?
(b) What is the half-life of the substances?

(a) 1/3=e^(-k5) Why am I wrong?
(b) I have no idea...

Answer 1

Answer:
(a) Ao (2/3)^(t/5)
(b) 8.5 years

Answer 2

Well, its 2/3 because 1/3 decays every 5 years, which still means 2/3 is left. Why they have the exponent in terms of t I'm not sure either, though. I would figure itd just be (2/3)^(5k). The exponent isnt negative, though, it is positive unlike your answer.

For part b, though, we need to start off by taking the natural log of both sides. That way we can solve for k.

Answer 3

taking the natural log of (a)?

Answer 4

Answer:
(a) Ao (2/3)^(t/5) You are right.

Answer 5

why t/5?

Answer 6

Actually, now that I think of it, it should be (2/3) = e^(5k), lol. This is the equation you were trying to get.

Answer 7

I couldnt tell you why t/5. Thats written to assume that k = (1/5) But we havent solved for k yet, which is what we are going to do, lol.

Answer 8

but the equation for decay is N=No e^(-kt)

Answer 9

k should be 5?

Answer 10

Well in general, decay or growth in general, you start with Y = ce^(kt) (same stuff, I just know it by different letters and notation). So even though we know it's decay, I like to just put that formula down first. When we get k we can see that it's negative or not.

Answer 11

do you think the answer is wrong?

Answer 12

So kind of what you were going off of, we would have (2/3) = e^(5k) because at t = 5

Answer 13

Since this is what you were getting at with your guess on a, hopefully it makes snese why I put that.

Answer 14

(2/3) = e^(5k)
ln(2/3) = 5k
k=-0.081

Answer 15

Which proves your negative point because of decay :P

Answer 16

what about t? how to find the half-life?

Answer 17

Well, instead of having 2/3 =, we say 1/2 =. Except now we know k, so we can say
1/2 = e^(-.081t)

Answer 18

After one half life, whatever you have is goign to equal 1/2, so we can justput it like that and solve for t.

Answer 19

i get 8.6 years but the answer is 8.5 years

Answer 20

why 1/2 ?

Answer 21

By definition, a half-life is the amount of time it takes for 50% of what you have to decay. We were told at the beginning that 1/3 decays after 5 years. So that's what we put, 2/3 remaining = e^(5k). Well, the time it takes for one half life leaves us with 1/2. If you dont know where to start with these problems, always know that 1/2 = e^(k*half-life)

Answer 22

The reason you got 8.6 is just rounding errors, thats natural.

Answer 23

oh right! thanks a lot i think i get it

Answer 24

Yep ^_^ Oh, and one more thing!

Answer 25

you are so kind!! thanks!

Answer 26

It's a fast tip, but let's say you have a problem where you're given the half-life. Like for example the problem says that the half-life of some element is 5,000 years and wants to know how much is left after 275 years. Well, if you know the half-life then this is ALWAYS true:
\[k=\frac{ -\ln2 }{ half-life }\]

Answer 27

Thanks for the tip!

Answer 28

but where does ln2 comes from?

Answer 29

\[-\ln2 = \ln(\frac{ 1 }{ 2 })\] and remember, when we were finding the half-life we set the equation = to 1/2. So it's just natural that the ln(1/2) or -ln2 would be int he formula. Everytime you do a problem, you can see for yourself that you would get the same for your k. It's just a thing to rememberinstead of solving for it every time.

Answer 30

okay.. I see. Thanks!

Answer 31

Yeah, sure. Let's see what wolf has left to say, though, lol.

Answer 32

Half life = (time * log 2) / log (beginning amount / ending amount)
If 1/3 decays every five years then we can set up the problem as
you start with 100 grams & 5 years later you have66.67% of it
Half life = (5 years * .3010) / log (100/66.67)
Half-life = (1.505) / (0.1760912591)
Half-life = 8.5467047488 years
Hey here's what I have to say

Answer 33

Sorry for not sticking to the traditional 'e' and 'ln' type of half-life calculations.
I like to stick with plain old English.

Answer 34

Lol, people put log andit confuses me xD I figure if ya mean natural log/ln, it should be moreobvious x_x log alone is ambiguous.

Answer 35

More than just using natural logs, I don't even like the traditional formulas that are used:
such as N=No e^(-kt)
Gee, that practically screams half-life doesn't it?

Answer 36

Im not used to even seeing it that way. I just know it as y = ce^(kt). Probably because I think of it more as a differential equation result.

Answer 37

Okay, anyway I should get going - it is 6:13 a.m. here in Boston - I better get some sleep (couldn't resist a half-life calculation though).

Answer 38

Night

Answer 39

Okay good night everyone - or is that good morning?

Answer 40

Thank you!

Answer 41

Sweet dreams!

Answer 42

wolf, would you mind explaining your equation Half life = (time * log 2) / log (beginning amount / ending amount)?
i think half-life = -ln2/lambda is much more common.
why do you use log instead of ln?

Answer 43

I use log because WAY back when - before there were calculators - people would use logs (or a slide rule) for numerical calculations. Common logs (or base 10 logs) were very convenient to use as compared to natural logs. By the way, I think you can use any base of logs for calculations (base 10, natural logs or whatever else) as long as you are consistent.
For example, to find the 5th root of 57.
take the log of 57 = 1.7558748557
Divide this by 5 = 0.3511749711
Looking up the anti-log of 0.3511749711 (or 10^.3511749711) we see it equals 2.2447861342 which is the fifth root of 57.
To make sure, we get a calculator and input (2.2447834261)^5 which equals 57.
In my next posting, I'll explain how I got the half-life formula.

Answer 44

Here's how I worked out the half life formula.
Whenever a time period of one half-life has gone by, the amount of original substance is reduced to one half. So
Ending Amount = Beginning Amount / 2^ (number of half lives)

So if a substance has undergone 3 half lives, the amount remaining is:
Ending Amount = Beginning Amount / 2^ (number of half lives)
Beginning Amount is represented as 1 so we have
Ending Amount = 1 / 2^ (number of half lives)
Ending Amount = 1 / 2^3
Ending Amount = 1 / 8
So after 3 half-lives, you have one eighth of what you started with.

The number of half lives = (elapsed time / half-life) so the equation can be rewritten as:
Ending Amount = Beginning Amount / 2^ (elapsed time / half life)
we can rearrange that to
2^ (elapsed time / half life) = (Beginning Amount / Ending Amount)
then taking logs of both sides
(Elapsed time / half life) * log 2 = log (Beginning Amount / Ending Amount)
Solving for half life
half life = Elapsed time * log 2 / log (Beginning Amount / Ending Amount)

Basically, I did that because I wanted to see precisely how the formula worked.

Answer 45

Thanks a lot for the information!!!! Now I understand your method!!!! Did you derive it yourself?
Can I ask one more thing?
"For example, to find the 5th root of 57.
take the log of 57 = 1.7558748557
Divide this by 5 = 0.3511749711
Looking up the anti-log of 0.3511749711 (or 10^.3511749711) we see it equals 2.2447861342 which is the fifth root of 57."
how do you know this? what about other roots? can we find them by log?

Million thanks!

Answer 46

Yes I derived the half-life formula myself.
As far as finding roots using logs it can be done but (like a slide rule) you'll only get approximate roots. (Yes, in school I was taught how to do this. But I was in school a LONG time ago).

Answer 47

Thanks a lot for the explanations!!!!!!!!!!! :)