Correct statement about random process of radioactive decay

[songoku]

Homework Statement

A physics student reads in a textbook that radioactive decay is a random process. This means that, for a sample of a given radioactive isotope, we cannot tell
A) what the radioactive isotope will decay into.
B) when the sample will start to decay.
C) which radioactive nucleus will decay next.
D) which type of radiation will be emitted.

Relevant Equations

none

The correct answer is (C) but I don't understand why (B) is wrong. Isn't (B) also the meaning of random process? I though (B) and (C) are both correct.

Thanks

 

[anuttarasammyak]

A sample made of a lot of radioactive nuclei started to decay immediately after it was generated.

 

[songoku]

Thank you very much anuttarasammyak

 

[kuruman]

Choice (B) is a bit ambiguous. To go to extremes, Tellurium-128 has a half-life of $7.7\times 10^{24}$ years via double beta decay. If I am given a sample, I "know" theoretically that it has started to decay but, arguably, the sample "has started to decay" as soon as I observe the first decay products which may or may not be in my lifetime.

 

[haruspex]

My objection to the question is that C is only part of what it means. So yes, you can arrive at C by elimination of the offered alternatives, but it also means there is no instant at which a decay can be predicted to occur. E.g., C would permit that a decay occurs exactly once every second, we just don’t know the order in which the atoms will decay.

It is also true, of course, that the likelihood at time t of a given undecayed atom decaying in the next interval of duration T depends only on T, not on t. Saying it is random is sometimes interpreted as implying that, but strictly it does not. The appropriate description of that is "memoryless".

 

[anuttarasammyak]

Choice (B) is a bit ambiguous. To go to extremes, Tellurium-128 has a half-life of 7.7×1024 years via double beta decay. If I am given a sample, I "know" theoretically that it has started to decay but, arguably, the sample "has started to decay" as soon as I observe the first decay products which may or may not be in my lifetime.

If I did it right, for 1 Bq radioactivity we should have $5 \times 10^7$ kg Tellurium sample, which is more or less weight of Titanic.

 

[gleem]

If you have ever monitored the decay of a radioactive isotope with a long half-life, you would note that the time between successive decays varies dramatically compared to the average. The decay rate is approximated by a Poisson distribution, which, for low decay rates, is extremely skewed toward larger rates (shorter time intervals between decays). You seem to get bursts of radiation between lulls. As the rate increases, the difference in time intervals between decays decreases and becomes more uniform.

The decay process of a single nucleus is like a kernel of popcorn; at the right time, it decays (pops). The use of the term start refers to the beginning of a process, but we only know the end when we detect the decay product at some distance from the nucleus. So B makes no sense.

 

[jbriggs444]

If you have ever monitored the decay of a radioactive isotope with a long half-life, you would note that the time between successive decays varies dramatically compared to the average. The decay rate is approximated by a Poisson distribution, which, for low decay rates, is extremely skewed toward larger rates (shorter time intervals between decays).

If you want the time between successive decays, you are concerned with an exponential distribution.

One does then have a sampling problem. Does one select a particular pair of successive decays by selecting a time at random or a decay at random? As I recall, there is a factor of two difference in expectation depending on that choice. If you select a time at random, the time between decays has an expected value twice what it would be if you had selected a decay at random.

[There is an expectation of the time until the next decay and a time from the previous decay. Both are characterized by the same exponential distribution. Sum them and you double the expectation]

The decay process of a single nucleus is like a kernel of popcorn; at the right time, it decays (pops). The use of the term start refers to the beginning of a process, but we only know the end when we detect the decay product at some distance from the nucleus. So B makes no sense.

I cringe a bit at comparing nuclear decay to popcorn popping. A popcorn kernel in the kettle is definitely going through a process. It is heating up until the internal pressure exceeds the bursting pressure of its shell. This takes time. The kernel is changing its state during this time.

By contrast, a nuclear decay is stateless/memoryless [as far as I know and as far as it is presented in the undergraduate physics classroom]. The decaying entity is not changing state as the supposed process progresses. This makes it hard for me to regard it as a "process".

 

[gleem]

I cringe a bit at comparing nuclear decay to popcorn popping.

Yeah, I kinda did too, but focusing on the commonality of the two, only monitoring the actual end of the processes, they are comparable.

 

[gleem]

If you want the time between successive decays, you are concerned with an exponential distribution.

I think some get the impression that the actual difference between successive decays is relatively constant and for high decay rates that is essentially true, but for very low count rates, it is very noticeably not true and an important consideration when performing a survey.

 

[jbriggs444]

I think some get the impression that the actual difference between successive decays is relatively constant and for high decay rates that is essentially true, but for very low count rates, it is very noticeably not true and an important consideration when performing a survey.

An exponential distribution is an exponential distribution. The only difference is scale.

I suspect that you are comparing the elapsed time between tens of decays (low decay rate) with the elapsed time between thousands of decays (high decay rate).

Imagine that one had a meter that counted detections and that this had a tunable sensitivity. So that it makes one audible tick for a tunable number of detections. Suppose that one tuned the sensitivity to produce one tick per second on average. With a low decay rate source and a high sensitivity, one would hear an irregular tick rate. With a high decay rate source and a low sensitivity, one would hear a regular tick rate.

Is that what you are getting at?

 

[gleem]

Imagine that one had a meter that counted detections and that this had a tunable sensitivity. So that it makes one audible tick for a tunable number of detections. Suppose that one tuned the sensitivity to produce one tick per second on average. With a low decay rate source and a high sensitivity, one would hear an irregular tick rate. With a high decay rate source and a low sensitivity, one would hear a regular tick rate.

Yes. The sensitivity and efficiency are independent of the decay rate, which is responsible for the variability in the rate of the radiation arrival at the detector. Geiger counters are particularly susceptible to large fluctuations in their reading at low decay rates.

 

[DEvens]

It's a multiple choice question. Usually you are advised to look for the "most correct" answer. Ordinarily multiple choice are for cases where the number of students is very large and so the instructor does not have time to mark long-form questions.

So if the answers are a bit, well, frayed around the edges, you are advised to hold your nose and move on. If there is no answer that could reasonably be called true then you might object to your instructor.