Posted by Lena on Saturday, February 19, 2011 at 5:56pm.
I got your answer as well.
Teachers do make mistakes.
I make about 2 per day, but my wife tells me I make more than that.
Aha thank you very much! :)
The formula is ln and not log AND it is much clear if you take a different approach. k = 0.693/t_{1/2} = 0.693/1.25E9 = 5.44E-10 yr^-1.
Suppose we are dealing with a total of 100 atoms, then we let x = K atoms that decayed to Ar and 100-x = atoms K remaining.
(Ar/K) = (x/100-x) = 3/1
Solve for x = 75 = Ar; 100-x = 25 = K remaining. Take 100 atoms.
ln(No/N) = kt
No = 100
N = 25
k = from above.
solve for t and I get something like 2.55E9 yr.
Check my work. Check my thinking.
Whats In?
I'm so confused :(
We were told to figure this out using half-life.
Our prof didn't even use logs or anything of that sort.
I think I know what I did wrong. If someone could let me know if my new understanding is right. If the ratio is 1:3 then you would have 1/4 K and 3/4 Ar so equation should have been:
1/4 = 1/2 ^(t/(1.25 *10^9))
Makes sense?
Then as an answer you would get:
(log 1/4 / log 1/2 ) * (1.25 * 10^9) = t
2.5 billion yrs = t
I think this is right! What do you think?
First, let me correct a typo I made and it affects the answer but only slightly.
k = 0.693/t_{1/2} = 0.693/1.25E9 = 5.54E-10
Then go through the x and 100-x bit I did earlier and x = 75 with 100-x = 25. Finally,
ln(No/N) = kt
ln(100/25) = 5.54E-10(t)
ln(4) = 5.54E-10 t
1.386 = 5.54E-10t and
t = 2.5E9 years which is slightly different because I typed the k wrong as 5.44E9 and not 5.54E9 the first time.
What is ln? It is the natural log (that's base e) as opposed to the log (base 10).
So you want to use half-life. I used half life when I calculated k above; however, there is another approach, which you came close to using that uses just half life and the ratio. That is
2^{n} = No/N. That is, 2 raised to the n power of half lives equals No/N which is the ratio of what you started with (No) and what you ended with (N). The error you made is in counting No/N as 3/1, which does, in fact, give you 1.58 as n or 1.58 half lives which equates to the age of about 2E9 years which you calculated. But No/N is not 3/1, it is 4/1. WHY? Go through the x and 100-x above again and you will see why but here is a summary. If you start with 100 atoms total K-40 at the beginning, then 75 of them decay to Ar-40 at the end of the time period and 100-x = 25 K-40 atoms are left. So the ratio of Ar/K = 75/25 = 3/1 (which is what the problem stated) BUT the ratio of the original K atoms (100) to the K atoms remaining is 25 or No/N = 4. So what do you do for this?
2^{n} = 100/25 = 4
2^{n} = 4
Of course you can look at this and see that n must be 2 because 2^2 = 4 OR you can do it with logs as I demonstrate when the problem is not so easily solved in your head.
2^{n} = 4
n*log 2 = log 4
n*(0.301) = 0.602
n = 0.602/0.301 = 2
So it has gone through 2 half lives.
1.25E9 years for the first half life.
2*1.25E9 = 2.5E9 for the second half life so it must be 2.5E9 years old.
I hope this helps. Isn't chemistry great!!
You typed in your correction while I was typing (with much much more explanation) my response. Yes you are exactly right. Good work figuring it out yourself. It's always so much more satisfying when you find the error yourself. Keep up the good work.
Thanks a bunch DrBob! :D
Sorry to re-enter the discussion at this point, but I just looked at the question again.
It shows the ratio 1:3, so as Lena also noted the equation should then have been
1 = 4(1/2)^(t/1.25)
1/4 = (1/2)^(t/1.25)
whether we take ln or logs does not matter
we either get
t/1.25 = ln(.25)/ln(.5) or t/1.25 = log(.25)/log(.5)
they both come out to
t/1.25 = 2
t = 2.5 , as Lena's teacher announced.
t/1.25 = ln(.25)/
Thanks Reiny :)!
But it does make a difference if we solve the problem as ln(No/N) = kt as I first responded. Log(No/N) = kt won't work, or at last I don't think it will.
I don't think log will work either because I tried it and I got a different answer :(
No Calculator Necessary.
Age 0: 0 Ar40
Age 1.25 billion years: half of the K40 is now Ar40, so 1 part of K40 and 1 part of Ar40
Age of 2.5 billion years: one quarter K40 and three quarters Ar40; so 1 part K40 to 3 parts of Ar 40