PHYS 620 Classical Mechanics II - University of Delaware. Fall 2018 Homework 9 Solutions.
PHYS 620 Classical Mechanics II - University of Delaware. Fall 2018 Homework 9 Solutions.PHYS620 Fall 2018 Homework 9 Solutions. 1. The cross section for scattering a certain nuclear particle by a co pper nucleus is 2.0 barns (if needed, look up the definition of a barn). If 109 of these particles are fired through a copper foil of thickness 10 μm, how many particles are scattered? (Copper’s density is 8.9 g/cm3 and its atomic mass is 63.5). 2. The differential scattering cross section for 6.5 MeV alpha particles at 120° off a silver nucleus is about 0.5 barns/sr. If a total of 1010 alphas impinge on a silver foil of thickness 1 μm and if we detect the scattered particles using a counter of area 0.1 mm2 at 120° and 1 cm from the target, about how many scattered alphas should we expect to count? (Silver has density 10.5 g/cm3 and atomic mass of 108.) 3. One can set up a two dimensional scattering theory, which could be applied to puck projectiles sliding on an ice rink and colliding with various target obstacles. The cross section σ would be the effective width of a target, and the differential cross section dσ/dθ would be the number of projectiles scattered in the angle dθ. (a) Show that the two dimensional analog of sin θ θ d d b db d σ = Ω is . d db d d σ θ θ = (b) Now consider the scattering of a small projectile off a “hard” sphere of radius R pinned down to the ice. Find the differential scattering cross section. (c) By integrating the differential cross section, show that the total cross section is 2R as expected. 14. One of the first observations that suggested his nuclear model of the atom to Rutherford was that several alpha particles got scattered by metal foils into the backward hemisphere, π/2 ≤ θ ≤ π – an observation that was impossible to explain on the basis of α b
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phys 620 classical mechanics ii university of delaware fall 2018 homework 9 solutions