Re: Nuclear rate of decay quandary

["Donald E. Simanek" <dsimanek@EAGLE.LHUP.EDU>]

On Thu, 22 Apr 1999, Kismet Talaat wrote:

> 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!

                                -(lambda)t
The equation to use is N  = N  e
                        f    o

Where lambda is the decay constant. The decay constant relates to the
half-life by

       ln(2)
T    = ------
 1/2   lambda

The solution above used the decay constant in one place and then took the
same value as lambda. That seems to be the blunder.

There's no need to obtain the half-life to work out this problem, however.
Applying the equation to the data 1024 and 1 we get lambda = -0.34657..
Then apply the equation to the interval of six minutes and you get 128
remaining, or 896 decayed.

> 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.

(delta N) = (lambda)No(delta-t) comes from the calculus equation for the
law of decay, dN = (lambda)No(dt), and cannot be applied to long intervals
of time, only very small ones. This equation should not be used for long
intervals of time.

> 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.

This is correct, with the given data.

> The question is....why are the first two solutions incorrect?

Noted above. However, this problem should be a candidate for the physics
problem hall of shame. During the last half-life represented by the data
only two atoms decayed. During the next half-life the one remaining is
supposed to decay. But due to the inherent statistical uncertainty of the
decay process itself, one atom may decay very promptly, or live nearly
forever before decaying. In short, the uncertainty in that last value "1"
in the data is *huge*, and makes any "estimate" based on it nearly
worthless in the real world of experimentation.

It would be an interesting problem for the rest of you to tackle. Based on
this data, estimate the uncertainty in the determination of half-life.
Also estimate the uncertainty of that answer of 128 atoms left after 6
minutes. Assume the uncertainties in the instrumentation, including the
times, are negligible.