Hi!
Thanks for your question.
Before I start, I'll define a few symbols to make it easier for you to
read:
No = N sub zero
h = lambda
* = multiplied by
/ = divided by
e(-h*t) = e to the power of lambda multiplied by t
ln = natural log
T1/2 = T sub 1/2
Now for the question:
a)
rearranging the formula to get a formula for t in terms of h/R
R = No*e(-h*t)/(No-No*e(-h*t)) the original formula
R = No*e(-h*t)/(No(1-e(-h*t)) factorising out No
R = e(-h*t)/(1-e(-h*t)) cancelling No from the top and
bottom of the fraction
R(1-e(-h*t)) = e(-h*t) rearranging
R - R*e(-h*t) = e(-h*t) expanding
R = e(-h*t) + R*e(-h*t) collecting like terms
R = e(-h*t)*(1+R) factorising e(-h*t)
R/(1+R) = e(-h*t) dividing both sides by (1+R)
e(-h*t) = R/(1+R) swapping the sides of the equation
ln(e(-h*t)) = ln(R/(1+R)) take natural log of both sides
-h*t = ln(R/(1+R)) result of taking the natural log
t = (-1/h)*ln(R/(1+R)) divide both sides by -h
therefore t in terms of lambda and R is:
t = (-1/h)*ln(R/(1+R))
b)
The well known equation relating lambda to half life T sub one half:
T1/2 = ln(2)/h
rearranging the formula found in a) to this form,
t = (-1)*(ln(R/(1+R)))/h rearranging
noting that in logs, -1*ln(x) = ln(1/x)
so t = (ln((1+R)/R))/h
this is in the same form as the well known equation where R/(1+R) = 2
and t = T1/2.
c)
using our formula in b),
for R = 0.5:
t = ln(3)/h
for R = 1:
t = ln(2)/h this is the half life!
for R = 2:
t = ln(1.5)/h
h is unknown since to determine lambda, we need the value of the half
life (or another variable)
For this reason I'll leave the answers as they are so it will be
easier for you to read and understand them (I won't substitute values
for ln(2) etc).
Any questions or problems, please request a clarification.
Search strategy:
Used google with the search terms "radioactive" "decay" and "solve for
t"
Useful Links:
Oklahoma State University:
A great site with lots of example problems
http://physics.okstate.edu/kaeding/physics1214/hw14.html
College of William and Mary:
This page shows the relationship between Pb and U in a graphical form.
http://www.wm.edu/CAS/GEOLOGY/geo101/Dating.html
Thanks,
Jenjerina-ga |
Clarification of Answer by
jenjerina-ga
on
05 Sep 2002 17:53 PDT
Hi,
Thanks for your clarification.
Since the formula we have derived in b) is:
t = (ln((1+R)/R))/h ----- (1)
and the general formula for the age of a rock is:
t = (ln(1+D/N))/h ---- (2) where:
D = the number of daughter atoms present today and
N = the number of parents atoms present today
ie, our ratio R = N/D
(for proof of this formula please see the geology page of Bryn Mawr
college:
http://www.brynmawr.edu/Acads/Geo/Geo101/geologic-time.html)
Rearranging (1+D/N) in formula number (2) gives:
(N+D)/N making one fraction by multiplying 1 by N/N
(N+D)/N ------ (3)
Substituting R = N/D into the (1+R)/R of formula number (1) gives
(1+(N/D))/(N/D)
(D+N)/N multiplying top and bottom by D
(N+D)/N rearranging
(N+D)/N ------ (4)
ie, formula (4) is the same as formula (3) therefore our formula (1)
is the same as the general formula (2).
Therefore, we can assume that Pb(208) is the stable daughter of
U(238). (Please see the Georgia Perimeter College geology page
http://www.dc.peachnet.edu/~pgore/geology/geo102/radio.htm for why I
was uncertain that Pb(208) was the stable daughter of U(238). The
website states that Pb(206) is the stable daughter of U(238) and that
Pb(208) is actually the stable daughter of Thorium 232).
So back to the question, since we can assume that Pb(208) is the
stable daughter of U(238) we can just substitue U(238)'s lambda into
formula (1). This is because, the rate of radioactive decay of U(238)
is the same as the rate of increase in Pb(208).
Therefore:
the from the 1/2 life formula
t = (ln(2))/h
h = (ln(2))/t rearranging
h = (ln(2))/4.5x10^9 substituting U(238) half life of 4.5x10^9y
h = 1.54*10^(-10)
so for R = 0.5:
t = ln(3)/1.54*10^(-10)
t = 7.13 * 10^9 years
for R = 1:
t = ln(2)/1.54*10^(-10)
t = 4.5 * 10^9 years
for R = 2:
t = ln(1.5)/h
t = 2.6 * 10^9 years
Please check my calculations because I did not have a scientific
calculator handy and had to use an online one
http://www.scientificcalculator.com/
I hope this clarifies your answer - if not please re-clarify.
Sure if you have any more different questions feel free to ask
Cheers,
Jenjerina
Search terms used:
3 Seaches on google:
"intervening decay OR decays radioactive age"
"parent daughter formula decay OR decays age ratio"
"Uranium 238 decay OR decays age ratio lead 208 example"
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