1. Sep 2, 2010

In lab, my partner and I are determining muon lifetime with a scintillator. I understand that the scintillator will help us find the average lifetime by measuring the difference between the time that a particle enters the scintillator and when in decays, but I am not able to figure out why this helps us find out the lifetime. All online sources I have found simply state this as a fact without going into the reasoning.

2. Sep 2, 2010

### Bob S

Your scintillator array needs to verify that a muon has entered the sensitive volume and has stopped. This usually requires at least 3 scintillators; Two scintillators, separated by ~10 cm or more of lead, copper, or other nonmagnetic high density material, in coincidence (10 nanoseconds or better) to determine the arrival of a minimum-ionizing particle. The coincidence circuit also needs a third scintillator in prompt anti-coincidence to verify that the stopping muon did indeed stop, and not exit the other side. All scintillator signals should be processed through a discriminator circuit so all minimum ionizing particles will produce a fixed amplitude output signal (NIM level (preferred) or a TTL signal). These three scintillators determine a "START" signal, which if possible is used to start a MCA (multichannel analyzer). The muon (probably a positive muon, more later) will decay with a positron with maximum energy of about 52 MeV. (See Michel rho parameter energy specrum in Fig. 1 in

http://arxiv.org/pdf/hep-ex/0311040v2

A fourth scintillator (maybe 10 by 10 by 20 cm) should be the sensitive volume in which the muon stops, and the mu-decay signal from this scintillator is used to generate a delayed STOP for the MCA. At first, the MCA oscillator should be set to advance the channel number count at ~50 nanoseconds per channel. So a 200-channel MCA would record about 10 microseconds (~5 lifetimes) of delay time.

Both positive and negative muons will stop in the stopping volume. Negative muons will form muonic atoms and likely form muonic (atomic) atoms and get captured by the nuclei in the stopping volume. These muons will give an anomalously short decay time (under 1 microsecond, if stopping volume material is over Z ~26).

The discriminator level on the stopping volume scintillator should be set to detect all electrons over ~ 10 MeV (See Fig. 1 above).

There will be a lot of background counts in all scintillators, especially the one used for the stopping volume. This whole setup will need to be well shielded with non-radioactive-contaminated high density material (lead or copper). Ideally, the background counting rate in the stopping volume scintillator is less that 1000 random singles counts per second (This is the most difficult shielding problem). This will assure that only ~1% of the STOP signals will be due to random background.

You need to be especially careful to properly set the discriminator thresholds properly. For a 10-cm thick stopping volume discriminator, it should be able to detect all non-stopping minimum ionizing muons (~ 20 MeV signal) when the prompt anti-coincidence signal is put into prompt coincidence with #s 1 and 2..

A simpler setup is to use the stopping-volume scintillator for both the MCA START and STOP signal. You will still need to set the discriminator sensitivity threshold with minimum ionizing particles (like cosmic rays).

What scintillators, phototubes, fast electronics, and MCA do you have?

Good luck. I hope this helps.

Bob S

Last edited: Sep 2, 2010
3. Oct 16, 2010

### Xia Ligang

Here you have to understand the process of muons' decaying. Muon has a meanlifetime. It can decay at any time, which is a poisson process. With some calculation, you will find that the time interval between a muon stopping and its decaying abide by an exponential distribution. Fitting the distribution, and you will measure muon's lifetime. I hope it will help you. I used to do this experiment in our lab. It took me many days and nights. Good luck to you!

4. Oct 16, 2010

### Bob S

If you do not determine the charge sign of the stopping muon, you will need to include in your analysis the decay rate of positive muons, with a decay rate of about 4.56 x 105 sec-1, and the capture rate of negative muons captured in muonic carbon atoms, with an estimated capture rate of 3.76 x 104 sec-1. See for example Equations (3.1) through (3.6) in

http://www.cs.cmu.edu/~byl/publications/phys410 lab2 report.pdf

Bob S

Last edited: Oct 16, 2010
5. Oct 16, 2010

Staff Emeritus
Let's try and focus, please? The OP's problem is clearly not that he has positive and negative muons, and it is highly unlikely that his scintillator is made out of iron. (Z=26)

6. Oct 16, 2010

### Bob S

The OP does not state where the muons come from. They do state that they are being stopped in scintillator (mostly carbon). Unless they produce the muons in an accelerator and have a magnetic focusing channel, the ratio of positive to negative muons is nearly one. In cosmic rays it is ~1.27:1. See

http://www.cs.cmu.edu/~byl/publications/phys410 lab2 report.pdf

Furthermore, in scintillator the negative muons form muonic (atomic) carbon atoms (the main component of scintillator). the capture rate in carbon, stated in the above URL and in the above post #4, is ~3.76 x 104 sec-1. This will reduce the measured lifetime.

Here is a quote from

http://www.fisica.unlp.edu.ar/~veiga/experiments.html [Broken]

"[Muon] Charge ratio can be estimated to be µ+/µ- = 1.2 (45% µ- and 55% µ+) and lifetime due to nuclear capture in carbon is aprox 1.93 μs (0.142 μs in iron). And we will measure both: 2.2 μs x 0.55 + 1.9 μs x 0.45 = 2.065 μs."

The 1.93 μs stated lifetime for µ- in carbon yields 6.2 x 104 sec-1 capture rate for µ- in carbon.

Very clearly, the capture rate stated in the post #4 represents the capture rate in carbon, not iron.

Bob S

Last edited by a moderator: May 5, 2017