The problem with both (1) and (2) comes from incorrectly using the
formula delta-N = lambda * No * t.
For finite t, the number of nuclei remaining does not stay at No
throughout the time interval. Therefore this equation must be
integrated to yield:
N(remaining) = No*exp(-lambda*t)
If you put your 6 minutes (t=360 seconds) into this exponential decay
equation with No = 1024 and lambda = 5.78e-3 then you get N(remaining)
= 128 which means N(decayed) is 1024-128 = 896. This is the
mathematically correct result for this problem.
The reason I said "mathematical result" is because it is not
necessarily the physical result. The whole nuclear decay and half-life
picture is only valid when there are sufficient numbers of nuclei to
make the statistical fluctuations minor compared to the overall numbers
of particles. Once you get down to small numbers (as being used here)
the half-life model is not statistically valid. We cannot have a
half-life of 2 minutes and start with 8 nuclei and expect to find 4
left after 2 minutes, 2 left after 4 minutes, 1 left after 6 minutes,
and... one-half of a nucleus left after 8 minutes... etc.? There will
be lots of statistical fluctuation in the early measurements, and, of
course, in the end we can't have less than one nucleus left. So the
model clearly breaks down once dividing by two yields partial nuclei...
and it actually gives pretty lousy results long before that.
Using small numbers like this is okay, I suppose, for getting students
used to the half-life idea. But to actually do the experiment and
expect good results, you need larger numbers.
Michael D. Edmiston, Ph.D. Phone/voice-mail: 419-358-3270
Professor of Chemistry & Physics FAX: 419-358-3323
Chairman, Science Department E-Mail edmiston@bluffton.edu
Bluffton College
280 West College Avenue
Bluffton, OH 45817
-----Original Message-----
From: Kismet Talaat [SMTP:Kismet_Talaat@FC.MCPS.K12.MD.US]
Sent: Thursday, April 22, 1999 11:18 AM
To: PHYS-L@LISTS.NAU.EDU
Subject: Nuclear rate of decay quandary
Hello all. I'm hoping someone out there can point out the error in
these two conflicting solutions to the same problem.
The problem states that ... "In a certain collection of nuclei there
were initially 1024 nuclei and 20 minutes later there was only one
nuclei left, the others having decayed. On the basis of this
information, how many nuclei would you estimate decayed in the first
six minutes?"
Several solutions were presented by the students:
1. Using Nf = No x 0.5^n (where n = # half lives), it was found that
10 halflives had passed for the sample to go from 1024 nuclei to one
nuclei. This was used to solve for the half life being 2 minutes, and
made the decay constant lambda = ln 2 / 120 seconds, or 5.78^-3/sec.
This decay constant was used with the formula for rate change ... delta
N = lambda x No x t .... (No = 1024, t = 360 seconds) and gave an
answer for the estimate of the number of nuclei that decayed (delta N)
in the first 6 minutes of 2129 nuclei that decay...more than the
original sample population!
2. A second solution offered was to use... delta N = lambda x No x
t.... with delta N = (1024 - 1) and t = 1200 sec (20 min).
This gave a decay constant of 8.325^-4 / sec. This different decay
constant would obviously yield the very different answer for delta N at
6 minutes of 307 nuclei....unreasonable because not even one half life
would have passed to obtain this number.
3. The third solution took the half life of two minutes, found the
number of half lives in 6 minutes to be n = 3, and used
Nf = No x 0.5^3 to give Nf at six minutes to be 128 nuclei left. This
gave a delta N of (1024 - 128) or 896 nuclei decayed.
The question is....why are the first two solutions incorrect?