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p226_ps1-SOLUTIONS-1

# p226_ps1-SOLUTIONS-1 - Physics 226 Problem Set#1 Due in...

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Physics 226: Problem Set #1 Due in Class on Thurs Sept 10, 2009 Some of the problems below require numerical calculations (integrals, random number generation, etc). You are free to use your favorite mathematical software package to do the problems. If you don’t have a favorite package, consider using ROOT. In doing these problems, you will need to know certain physical constants (particle lifetimes, radiation lengths of specific materials, etc). You can find everything you need on the Particle Data Group (PDG) web site: http://www-pdg.lbl.gov/ Take some time to learn to navigate this site. We’ll be using it extensively. 1. Particle ID using a time-of-flight detector The CDF experiment has a time-of-flight (TOF) system. The purpose of this detector is to distinguish π ± from K ± . The system has a time resolution σ t = 100 ps and is located 140 cm from the point where the π and K are created. (a) Let’s begin by deriving an expression for the difference in flight time for two relativistic particles of masses m 1 and m 2 with the same momentum p that travel a distance . (Hint: start with the relativistic expression β = p/E and Taylor expand the energy E for the case where m << p ). For ultrarelativist particles ( m 1 , m 2 << p ) the time of flight is t = v = c E pc where E is the energy and p is the momentum of the par- ticle. Thus for two particles with the same momentum and different masses Δ t = t 1 - t 2 = c q ( pc ) 2 + ( m 1 c ) 2 pc - q ( pc ) 2 + ( m 2 c 2 ) 2 pc Taylor expanding q ( pc ) 2 + ( mc 2 ) 2 = pc q 1 + ( mc 2 ) 2 / ( pc ) 2 = p (1+ 1 2 ( mc ) 2 / ( pc ) 2 + ... ) we find Δ t = 2 c ( m 1 c 2 ) 2 - ( m 2 c 2 ) 2 ( pc ) 2 1

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setting c = we get: Δ t = 2 ( m 1 ) 2 - ( m 2 ) 2 p 2 (b) Using the expression you have derived, find the maximum momentum for which π and K can be separated at better than the 90% confidence level. (If you are not familiar with the concept of a “confidence level” see the review of statistics in the PDG.) You may assume all errors are Gaussianly distributed. A 90% confidence level separation corresponds to 1.64 σ for a two-sided cut. However, since we are only cutting on one side of the Gaussian, we should ask that 10% of the events be in one tail (so 20% in both tails. That corresponds to 1.28 σ from each peak, or 2.56 σ between the peaks. So we are looking for the momentum where the time difference Δ t = 2 . 56 × 100 ps which means: ( pc ) 2 = 2 c ( m 1 c 2 ) 2 - ( m 2 c 2 ) 2 Δ t ( pc ) = v u u t 1 . 40 m 6 × 10 8 m/s (0 . 494 2 - 0 . 140 2 )GeV 2 258 × 10 - 12 s p = 1 . 43GeV / c (c) Suppose a factor of 3 more pions than kaons are produced at a mo- mentum of 1.5 GeV. What cut on time-of-flight must be applied in order to produce a sample of candidate kaons with a purity p of 80% ( p = n K / ( n K + n π )? What is the efficiency ( = n passing K /n total K ) of this cut for kaons at this momentum? First, we find the separation in time of arrival of the π and K : Δ t = 1 . 4 6 × 10 8 (0 . 494 2 - 0 . 140 2 )GeV 2 (1 . 5GeV / c) 2 = 2 . 33 × 10 - 10 sec = 233 ps Thus, if the mean time of arrival of the π is -233 ps, the mean time of arrival of the K is 0 ps. In each case, the 2
arrival time is Gaussianly distributed with a σ = 100 ps. If we call everything a K that arrives after time T , then the

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