# Coincidence events

Manojg
Hello,

I have a question about coincidence event. Let us take the decay of deuteron to proton and alpha particle, (d, pα). p and α goes in opposite direction. So, if we put two detectors in opposite directions, one of them will detect p and another will detect α simultaneously (within a small time window).

If the solid angle covered by both detectors are same (say Ω) and if one of the detector detect p then it is sure that another one will detect α. If N0 be the total decay rate then number of coincidence event detected will be N0Ω.

However, if the solid angle covered by the detectors are different, say Ω1 and Ω2 such that Ω1 > Ω2 then the number of coincidence event will be equal to the number of events detected by the second detector because other extra event detected by the larger detector won't be detected by the smaller detector.

Are these reasoning right? Because in a book, I saw the number of coincidence event is N0Ω1Ω2.

Thanks.

## Answers and Replies

Mentor
A deuteron is a stable nucleus with a single proton and a neutron. It cannot decay in the way you describe.
Lithium-5 quickly decays to proton+alpha particle - you have to produce it directly before the decay, and this usually will not happen without any momentum transfer.

If your decay is back to back:
If N0 be the total decay rate then number of coincidence event detected will be N0Ω.
Or 2 times this value, if both detectors are sensitive to both particles. And I guess you should divide it by the full solid angle of a sphere, or express Ω as ratio to that.

then the number of coincidence event will be equal to the number of events detected by the second detector because other extra event detected by the larger detector won't be detected by the smaller detector.
This is an upper limit.

Are these reasoning right? Because in a book, I saw the number of coincidence event is N0Ω1Ω2.
That would correspond to independently moving, isotropic decay products.