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UT Arlington PHYS 3446 - Measurement of Cross Sections

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PHYS 3446 – Lecture #4Assignment #2 Due Weds. Sep. 22Scattering Cross SectionTotal X-Section of Rutherford ScatteringMeasuring Cross SectionsSlide 6Slide 7Slide 8Lab Frame and Center of Mass FrameLab Frame and CM FrameNow some simple arithmeticSlide 12Slide 13Relationship of variables in Lab and CMScattering angles in Lab and CMMonday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 1PHYS 3446 – Lecture #4Monday, Sep. 20 2010Dr. Brandt1. Differential Cross Section of Rutherford Scattering2. Measurement of Cross Sections3. Lab Frame and Center of Mass Frame4. Relativistic Variables***HW now due Weds***Labs due at next lab***Everyone should go to lab Friday 24 and Fri Oct. 1 weeks as we will have a half lecture before each labMonday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 2Assignment #2 Due Weds. Sep. 221. Plot the differential cross section of the Rutherford scattering as a function of the scattering angle  for three sensible choices of the lower limit of the angle. (use ZAu=79, Zhe=2, E=10keV).2. Compute the total cross section of the Rutherford scattering in unit of barns for your cut-off angles. 3. Find a plot of a cross section from a current HEP experiment, and write a few sentences about what is being measured.4. Book problem 1.10 *New Problem*•5 points extra credit if you are one of first 10 people to email an electronic version of a figure showing Rutherford 1/sin^4(x/2) angular dependenceMonday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 3Scattering Cross Section•For a central potential, measuring the yield as a function of , or differential cross section, is equivalent to measuring the entire effect of the scattering •So what is the physical meaning of the differential cross section?This is equivalent to the probability of certain process in a specific kinematic phase space•Cross sections are measured in the unit of barns:-24 21 barn = 10 cmMonday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 4Total X-Section of Rutherford Scattering•To obtain the total cross section of Rutherford scattering, one integrates the differential cross section over all :•What is the result of this integration?–Infinity!!•Does this make sense?–Yes•Why?–Since the Coulomb force’s range is infinite (particle with very large impact parameter still contributes to integral through very small scattering angle)•What would be the sensible thing to do?–Integrate to a cut-off angle since after certain distance the force is too weak to impact the scattering. (0>0); note this is sensible since alpha particles far away don’t even see charge of nucleus due to screening effects.Totals =( )02 sindddpsp q q q =W�22103' 18 sin4 2sin2ZZ edEqpq� �� �� �� �� �� ��Monday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 5Measuring Cross Sections•With the flux of N0 per unit area per second•Any  particles in range b to b+db will be scattered into  to  -d•The telescope aperture limits the measurable area to• How could they have increased the rate of measurement?–By constructing an annular telescope–By how much would it increase?TeleA =2/dsinRd R dq q f� =2R dWMonday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 6Measuring Cross Sections•Fraction of incident particles (N0) approaching the target in the small area =bddb at impact parameter b is dn/N0. –so dn particles scatter into R2d, the aperture of the telescope•This fraction is the same as –The sum of  over all N nuclear centers throughout the foil divided by the total area (S) of the foil.–Or, in other words, the probability for incident particles to enter within the N areas divided by the probability of hitting the foil. This ratio can be expressed as 0dnN- =Nbd dSf q( ),NSs q fD =Eq. 1.39Monday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 7Measuring Cross Sections•For a foil with thickness t, mass density  , atomic weight A:•Since from what we have learned previously•The number of  scattered into the detector angle () isN =A0: Avogadro’s number of atoms per moledn =0tSAAr0 0N tAAr( )0,dNN dS ds q fWW( ),ddds q fW=W0dnN- =Nbd dSf qEq. 1.40Monday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 8Measuring Cross Sectionsdn =•This is a general expression for any scattering process, independent of the theory•This gives an observed counts per second0 0N tAAr( )0,dNN dS ds q fWW( ),ddds q fW=WNumber of detected particles/secProjectile particle fluxDensity of the target particlesScattering cross sectionDetector acceptanceMonday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 9Lab Frame and Center of Mass Frame•So far, we have neglected the motion of target nuclei in Rutherford Scattering•In reality, they recoil as a result of scattering •This complication can best be handled using the Center of Mass frame under a central potential•This description is also useful for scattering experiments with two beams of particles (moving target)Monday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 10Lab Frame and CM Frame•The equations of motion can be written1 1m r =r&&2 2m r =r&&i� =rwhereSince the potential depends only on relative separation of the particles, we redefine new variables, relative coordinates & coordinate of CM1 2r r r= -r r rCMR =rand1 1 2 21 2m r m rm m++r r( )1 1 2V r r- � -rr r( )2 1 2V r r- � -rr rˆiirr�+�ˆii irqq�+�ˆsinii i irfq f��1, 2i =2rrMonday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 11Now some simple arithmetic•From the equations of motion, we obtain1 1 2 2m r m r- =r r&& &&1 1 2 21 2CMm r m rRm m+=+r r&& &&r&&( )1 2 1 2 m m r m r+ = �r r&& &&( )( )21 2 21 2mm m r m rm m- + =+r r&& &&•Since the momentum of the system is conserved:•Rearranging the terms, we obtain( )1 1 2 1m r m r r- - =r r r&& && &&( )1 2 1 2m m r m r- +r r&& &&( ) ( )1 21 2ˆ ˆ ˆ2r r V r r V rr r r� �� � �=- - =-� �� � �� �r r0 = �1 1 m r =r&&2 2m r- =r&&( )2 1m r r- -r r&& &&( )211 2 mr rm m=+r r&& &&( )1 21 2m mrm m=+r&&( )ˆr V rr�-�r( )1 21 22m mrm m+r&& �Monday, Sep. 20, 2010 PHYS 3446, Fall 2010 Andrew Brandt 12Lab Frame and CM Frame•From the equations in previous slides( )1 2 CMm m R+ =r&&and•What do we learn from this exercise?•For a central potential, the motion of the two particles can be decoupled when re-written in terms of –a relative coordinate–The coordinate of center of


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