exercise 13.2 lision is calculated from conservation of energy as 4.3 × 10-15 m
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exercise 13.2
lision is calculated from conservation of energy as 4.3 × 10-15 m or 4.3 fm. For elastic collisions, the kinematic relation between the scattering angle ? in the center-of-mass or relative coordinate frame and the scattering angle in the laboratory frame-of-reference is sin 8 tan 6 (13.10) where mp is the mass of the incident projectile and mr is the mass of the initially stationary target (Figure 13.4). For example, corresponding to e'-30 in the laboratory frame, the scattering angle in the center-of-mass system (as well as in the relative coordinate frame) is ?-34.2" The distance of closest approach corresponding to this scattering angle is 9.5 fm, which is close to the nuclear radius of aluminum, so that small departures from Rutherford scattering may occur. We then calculate the Rutherford differential cross section in the center-of-mass frame for this scattering angle as d 1.53 x 10-2 m2 1.53 x 10-24 cm2 1.53 barns Thisis the differential cross section per steradian. At this scatteri gangle one finds 286 Chapter 13 Scattering and, for the differential cross section in the laboratory frame, ?.desin. _ 195bams _ 195 x ir" ? dn sin 0' - AY If the solid angle subtended by the detector is d10-6 steradian, the calculated cross section for this outcome, is d?-1.95 >Explanation / Answer
Some information is not provided The intensity of the beam and the number of particles scattered are not given.
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