Question

Protons of energy 4.7 MeV are incident on a copper foil of thickness 4.9 x 10-6 m. What fraction of the incident protons is scattered at the following angles? (The density of copper is 8.9 g/cm", and its molar mass is 63.5 g/mol. (a) greater than 90° 0.000026 (b) less than 5°

Answer 1

Answer! Pagerto Given that, Proton's of Energy = 4.7 mer Coppear foil of thickness = 4.9 x10 m The density of copper is = 8.9 g/cm3 molar mass = 63.5g imol The foraction of incident particles scattered through an angle o is given by f = (Tib?) (net) where b= kq pQ cot 0/2 26K : The column density 4.9x106m net = mall density & Avagadros molar mals - thickness = (8.9 91cm3) (6.02 x 1023 atoms 14 ) moles (63.59m) (4.axio" cm) = 4:1301019 atoms/cm²

PageNo@ The impact Parameter for proton on Coppen nuclici is b = K[e) (age) cote 2x(uimer) b= 14.5 kez coton uilmer = 14.5 x 144 ev-mm 4.4 X ro@ev b=( 5.09x1093cm) Cofola Cs) Gosuaken on 26°. 0-90° coto - f = t (5.09x10-13 cm) x (4.13 x 10'9) f = 3,14 x 25.908x 4.13x107 If= 3.35x10-5

Page No@ Os go cot (9) = cot Ca's) = 22.903 f:7 (5.09x10=13*22.903)?x (4113x1019) = 3,14 x 13.59 x103 *4.13x107 f = 1176x102