Particle RANGE Through Material and Particle Lifetimes Dan Claes CROP Fall 2006 Workshop Saturday, October 14, 2006

EarthMoon While the size of the moon represented is roughly to scale for the earth as shown… …the distance separating them is about 12  too small!

EarthMoon

Scale of the atomic orbital is about a factor of 5,000  too small! The nucleus would be invisibly small if corrected in scale to the size of depicted orbital. Atoms really are mostly empty space!

V V Muon in head-on collision with a nucleus Muon in head-on collision with an electron

Actually if an incoming “sees” n nuclei per unit area “in its way” there will be Zn electrons per unit area!

V

Carbon 6 C Oxygen 8 O Aluminum 13 Al Iron 26 Fe Copper 29 Cu Lead 82 Pb What about a single, high energy, charged particle? While the mass of matter is due primarily to it’s nuclei The volume of matter is due primarily to it’s electron clouds

Imagine a narrow, well-collimated beam of mono-energetic particles passing through a slab of matter EoEo E EoEo

EoEo E Energy loss E1E1 EE

EoEo E Energy loss E1E1 EE If the target is thick, this implies that the overall mean energy loss  thickness For sufficiently high initial E 0 (or thin enough targets) all particles get through.

EoEo E 0 but if the target is thick enough thickness 

Occasionally might just catch a muon stopped within the the thickness of our detector! Such an occurrence would be “signaled” by a coincidence between the top two counters with NO SIGNAL in the bottom.

n  p + e   e    e   e +  Ne *  Ne +  N  C + e   e Pu  U +  Fundamental particle decays Nuclear decays Some observed decays

 + DECAY MODES  1  + 

Particle Data Group Created: 06/18/2002     0  150 mesons!!

It almost seems a self-evident statement: Any decay that’s possible will happen! What makes it possible? What sort of conditions must be satisfied? Total charge q conserved. J (angular momentum) conserved.

p n Particle Data Group Created: 06/18/2002

probability of decaying (at any time - now or later) = constant ???? What’s this mean equally likely at any instant ???? must be expressed as a probability per unit time If we observe one, isolated nucleus it is equally likely it decays this moment  t as any other moment  t (even years from now) It either decays or it doesn’t. A quantum mechanically model for this random behavior suggests

Suppose a given particle has a 0.01 probability of decaying in any given  sec. Does this mean if we wait 100  sec it will definitely have decayed? If we observe a large sample N of such particles, for 1  sec how many can we expect to have decayed? Even a tiny speck of material can include well over trillions and trillions of atoms! 0.01N

Imagine flipping a coin until you get a head. Is the probability of needing to just one flip the same as the probability of needing to flip 10 times? Probability of a head on your 1st try, P(1) = Probability of 1st head on your 2nd try, P(2) = Probability of 1st head on your 3rd try, P(3) = Probability of 1st head on your 10th try, P(10) = 1/2 1/4 1/8 (1/2) 10 = 1/1024

What is the total probability of ALL OCCURRENCES? P(1) + P(2) + P(3) + P(4) + P(5) + =1/2+ 1/4 + 1/8 + 1/16 + 1/32 +

A six-sided die is rolled repeatedly until it gives a 6. What is the probability that one roll is enough? 1/6 What is the probability that it will take exactly 2 rolls? (probability of miss,1st try)  (probability of hit)= What is the probability that exactly 3 rolls will be needed?

# decays  N (counted by a geiger counter) the size of the sample studied   t time interval of the measurement each decay represents a loss in the original number of radioactive particles fraction of particles lost Note: for 1 particle this must be interpreted as the probability of decaying. This argues that:  constant This is the decay constant

imagine the probability of decaying within any single second is p = 0.10 the probability of surviving instead during that same single second is P(1)= 0.10 = P(2)= 0.90  0.10 = P(3)=  0.10 = P(4)=  0.10 = P(5)=  0.10 = P(6)=  0.10 = P(7)=  0.10 = P(8)=  0.10 = P(9)=  0.10 = P(N) probability that it decays in the Nth second (but not the preceeding N-1 seconds) 1  p = 0.90

Probability of living to time t =N sec, but decaying in the next second (1-p) N p Probability of decaying instantly ( t =0) is? Probability of living forever ( t   ) is? 0 0

We can calculated an “average” lifetime from  (N sec)×P(N) (1 sec)×P(1)= (2 sec)×P(2)= (3 sec)×P(3)= (4 sec)×P(4)= (5 sec)×P(5)= N=1  sum= sum= sum= sum= sum= sum=

the probability of decaying within any single second p = 0.10 = 1/10 = 1/  where of course  is the average lifetime (which in this example was 10, remember?)

This exponential behavior can be summarized by the rules for our imagined sample of particles fraction still surviving by time t = e  t where = 1/  (and   is the average lifetime)

Number surviving Radioactive atoms time  log N

probability of surviving through to time t then decaying that moment (within t and  t ) or …and for the calculus savvy…

Some Backgrounds to this estimate: Accidentals: 2 stray cosmic rays, each passing individually through just one detector, but by sheer chance coincident, can set off the trigger. Another accidental passing through the 2nd counter within 10  sec, will fake a muon decay. accidentals rate  bottom counter’s singles rate × gate width Inefficiencies: Can trigger when a cosmic track is seen by the top two counters, but missed (since its not 100% efficient) by the bottom. If an accidental passes through the bottom within 10  sec it gets counted as a muon decay. 2-fold trgr rate(top 2) × (1  efficiency of bottom)  bottom’s singles rate × gate width