- #1
physguy09
- 19
- 0
I have been doing an experiment on muon lifetime, and am wondering if my count rate is correct.
A little about the experiment:
I have two scintillators (plastic) set up above a lead glass absorber, and a third scintillator(plastic), under the lead glass to veto muons that do not stop in the Pb-glass. If the Logic Unit (LU) detects coincidence, a start signal is sent to the Time-to-Amplitude converter. If the muon stops in the Pb-glass, a stop signal is sent. The area of the top portion of the lead glass (where the muon is incident) is roughly 225 cm2.
So assuming this is correct, I figured I should be getting:
10,000 [tex]\frac{muons}{meter^2 minute}[/tex] * ([tex]\frac{1 meter^2}{10,000 cm^2}[/tex])*230 cm2 = 230 muons/minute.
I have observed a count rate of about 50 muons/day, and this seems to be rather low. There are 10 floors above the location of my apparatus, so I know that there is some attenuation for that reason, but 50 muons/day is no where near the 331,000/day I should be getting.
A little about the experiment:
I have two scintillators (plastic) set up above a lead glass absorber, and a third scintillator(plastic), under the lead glass to veto muons that do not stop in the Pb-glass. If the Logic Unit (LU) detects coincidence, a start signal is sent to the Time-to-Amplitude converter. If the muon stops in the Pb-glass, a stop signal is sent. The area of the top portion of the lead glass (where the muon is incident) is roughly 225 cm2.
So assuming this is correct, I figured I should be getting:
10,000 [tex]\frac{muons}{meter^2 minute}[/tex] * ([tex]\frac{1 meter^2}{10,000 cm^2}[/tex])*230 cm2 = 230 muons/minute.
I have observed a count rate of about 50 muons/day, and this seems to be rather low. There are 10 floors above the location of my apparatus, so I know that there is some attenuation for that reason, but 50 muons/day is no where near the 331,000/day I should be getting.