I am stuck at part C. For A I have: 2.24 gamma factor for the Pions For B I have

Question

I am stuck at part C.

For A I have: 2.24 gamma factor for the Pions

For B I have: 5.8*10^-8 s in lab frame

Am I on the right track, can someone direct me in the right way with the rest of this problem please? Thanks.

A group of pi mesons (pions) are observed to be traveling at u' = 0.8c in a physics laboratory. The mean lifetime (let's call it A) for unstable particles undergoing exponential decay is the average time for a group of particles to be reduced to 1/e or their original number. We can express this mathematically as: Nt = N0 exp(-deltat/lambda). We see then that for deltat = lambda, Nt = N0 exp(-1) = (1/e) middot N0. What is the 7 factor for the pions? If the pions' proper mean lifetime is A = 2.6 times 10-8 s, what is the lifetime (lambda') as observed in the laboratory frame? If there were initially 32,000 pions, how many will be left after they have traveled 36 meters from the source to a detector (as measured in the laboratory frame)? Show that this number is the same, as calculated in the rest frame of the pions. What would the answer to (d) be if there were no time dilation (i.e.. if deltatau = deltat')? How long does the distance of 36 meters in the laboratory frame appear to the pions in their rest frame? Use the result from (f) and the proper time elapsed (as calculated in the rest frame of the pions) to find the velocity of the laboratory frame, as calculated in the rest frame of the pions.

Explanation / Answer

C. Time taken to travel the distance 36m is t = d / v

= (36m) / (0.80c)

= (36m) / (2.4 *108s) = 15 *10-8s

Now N = N0 e^-(0.693/T1/2) t

= 32000 e^(-0.693* 2.4*10^-8/2.6*10^-6)

= Now 168.78 pions left after travelling the distance 36 m.

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