The first question was only apparently easy - even too much, from the outset. The fact is, the devil is always hiding in the details, as I immediately realized as I tested it by asking an experienced colleague to answer it. He got part of the question wrong, but in doing so he clarified to me that there was a non trivial aspect below the surface.
I think it is instructive to discuss the question and its subtleties here, if you care to listen. So here it is. It is composed of three sub-questions.
1 - You have a sample containing a fixed amount - say a femtogram - of a unknown homogeneous radioactive substance, and you start the clock and detect its decays using a counter with zero background. After 1' you observe that 400 decays have been recorded. With that information can you estimate the lifetime of the substance ? Please provide a yes or no answer, and an explanation of your reasoning.
2 - (if the student correctly answered NO to the first question) You perform a second counting for the second minute, consecutive after the first. You record 100 decays. Can you now estimate the lifetime ? How do you do it ? We are not interested in the answer, only in the procedure.
3 - In light of the magnitude of the above two measurements, what would you give as an estimate of the standard deviation of the first count, i.e. 400 ?
Solution: part 1
So let us reason on the problem, starting with the first question above. A radioactive substance of unknown nature is given, so we do not know its atomic number. This makes the information we have on its mass (1 picogram) useless for the sake of determining the total number of active nuclei initially present in the sample as we start the clock and record decays with the counter. (On the right is shown a graph of an exponential curve describing the decay of a radioisotope, carbon 14, which is very commonly used to date old materials. Careful though - the graph is only qualitative; e.g. the units on the horizontal axis are not equally spaced, and those on the vertical one are also messed up).
What we may assume of the system is that a sample of N0 atoms decay with an exponential decay law, governed by a lifetime tau. The formula giving the number of undecayed particles at time t, given N0 particles at time t0=0, is simply
N(t) = N0 * exp (-t/tau).
So as you can see there are two unknowns in the system. However, we only have one measurement - the number of counts in the first minute. This should be sufficient to convince you that the provided information is insufficient to determine the lifetime. This is a proper answer to the first question.
Solution: part 2
As for the second question: you now have two measurements, and you should be able to make it. How ? Well, if the formula above determines the number of undecayed particles at time t, you may write the number of decays in the first minute as D(t) = N0 * [1 - exp(-t/tau)]. As t=1 minute, you may write
N1 = N0 * [ 1- exp(-1/tau) ]
where we prefer to keep N1=400 expressed as a parameter.
For the second minute, you need to subtract from the unknown N0 the number of decays occurred in the first minute. So you can write:
N2 = (N0-N1) * [1-exp(-1/tau)]
again choosing to keep N2=100 expressed literally. Note that in this second equation we still write "1/tau" in the exponential, as we are now counting time from the first minute to the second minute.
The two equations above may be solved effectively if we first operate a simple change of variable: setting x = exp(-1/tau), we obtain
N1 = N0 * (1-x)
N2 = (N0-N1) * (1-x)
Now let us substitute N0, as expressed in the first equation, into the second one. We obtain
N2 = [N1/(1-x) - N1] * (1-x)
which simplifies into
N2 = N1 - N1 *(1-x)
that is (N1-N2)/N1 = 1-x, or x=N2/N1.
Now if we remember that x=exp(-1/tau) we easily find:
N2/N1 = exp(-1/tau)
and taking the log
log(N2/N1) = -1/tau
so
tau = 1/log(N1/N2)
which is the estimate of the lifetime. In our case N1=400, N2=100 and
we get tau = 1/log(4) = 0.72 minutes.
Note that we solved the question directly, but the student would only be required to explain roughly how he would do it: saying "I think I can construct a system of two equations each of which expresses the number of decays in the corresponding minute as a function of initial number of nuclei and lifetime, and then solve for the lifetime" would be an excellent answer.
Solution: part 3
Okay, so far so good. Now for the tricky part. You should first of all know that events occurring at random and in independent fashion (like nuclear decays) from a process with a constant rate follow a Poisson distribution, meaning that if there is a certain average number of events that should take place in a given time interval, then the actual observed events distribute according to the formula
P (N | m) = exp(-m) m^N / N!
Here P represent the probability that I observe exactly N events, given that the expected rate was m.
Most physics students after obtaining a Master do know the Poisson distribution, and they will likely also know that the standard deviation of a number N drawn from a Poisson distribution is the square root of the mean m. So if they are asked to estimate what is the standard deviation on a number of Poisson event counts, they will be likely to say this is the square root of N. In our exercise sqrt(400)=20.
But is that the correct answer in this case ? No, it is not. A question one might ask to the student who reported 20 as the result of the third part is "are you perfectly sure that the distribution of the decays follows a Poisson distribution" ? For, you see, it does not! In fact, the Poisson requires that the mean m in the equation above be constant. But for a sample of nuclei undergoing decay, m varies, as the source gets increasingly depleted!
The issue becomes murkier once you decide to start discussing how precise can the number of nuclei in the sample be - as after all, if you produce 100 "equal" samples of the radioactive substance, you may well expect that some of them will contain a larger number of nuclei than the others. So the estimate of standard deviation on N1 will have to suffer from that variation too.
Or one could reason that, as we know nothing of the nuclear reaction going on in the sample, and we have not explained how the counter works, it is conceivable that a nucleus undergoes multiple decays in succession, each with its own lifetime, and each being counted by recording gamma rays or other decay products in our counter.
You see, physics is always difficult when you consider real-life situations; that is why it is much better to discuss idealized cases instead. But putting a student in the situation of having to consider a real-life situation is interesting, as the way he or she attacks the problem, takes approximations, makes assumptions, and draws conclusions or questions the results of calculation is very interesting and can be a very useful evaluation tool.
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