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Re: Nuclear rate of decay quandary



Dear Kismet,

The problem is that the rate equation is delta N/delta t = - lambda
* N. This is a differential expression that holds only for times that are
very short relative to the half life of 2 minutes.

You probably don't remember me, but I once met you when I was
serving as a rotator at NSF back in 1989.

Mark

Dr. Mark H. Shapiro

Chair, Department of Physics
California State University, Fullerton
P.O. Box 6886
Fullerton, CA 92834-6866

Phone: (714) 278-3884
FAX: (714) 278-5810
Cell/Pager (714) 350-3575
e-mail: mshapiro@fullerton.edu


-----Original Message-----
From: Kismet Talaat [SMTP:Kismet_Talaat@FC.MCPS.K12.MD.US]
Sent: Thursday, April 22, 1999 8:18 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?