Question

Problem 10.23 An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 9.9 MeV (9.9 x 100 eV), and heads in the +x direction straight toward a gold nucleus (containing 79 protons and 118 neutrons). The particles are initially far apart, and the gold nucleus is initially at rest. Answer the following questions about the collision. What is the initial momentum of the alpha particle? (You may assume its speed is small compared to the speed of light). ai = = > kg.m/s What is the initial momentum of the gold nucleus? P Au, i = < > kg.m/s What is the final momentum of the alpha particle, long after it interacts with the gold nucleus? > kg.m/s Paf = What is the final momentum of the gold nucleus, long after it interacts with the alpha particle? P auf = = > kg.m/s What is the final kinetic energy of the alpha particle? Ka = What is the final kinetic energy of the gold nucleus? KAus =

Answer 1

(a) The kinetic energy of the alpha particle is, Ka =(9.9x109 ev 1.602x10-19 =19.9x10e 1eV = 1.58598x10-12 = 1.6x10-12 J The initial velocity of the alpha particle is, m Vai = Ka 2K ai ma 2(1.59598x10-12) 2m, + 2m, 2(1.59598x10-12) 12(1.67x10-2) + 2(1.67x10-2) = 2.18595x10 m/s =2.2 m/s The initial momentum of alpha particle is, Pari = mVia = 4(1.67x10-27)(2. 18595x107) = 1.46x10-19 kg.m/s (b) The initial momentum of gold nucles is equal to zerom beacause initially gold is at rest. Pawy = 0 kg.m/s (c) The final velocity of alpha particle is, (ma ma vai me + manai 4(1.67x10-2) – 1971.67x10-2) 5(4(1.67x10-2) + 197(1.67x10-27)|| |(2.18595*10') =-2.0989x107 =-2.1x107 m/s The final momentum f alpha particle is,

Pas = m Var = 4(1.67x10-27)(-2.0989x10) =-1.40x10-19 kg.m/s (d) The final velocity of gold nucles is, 2m Vai mo+mdn 2(4)(1.67x10-27) (2. 18595x10) (4+197)(1.67x102) = 8.70 x 10 m/s The final momentum of alpha is, Pas = m du Vert = (197)(1.67x10-2)(8.70x10) = 2.86x10-19 kg.m/s (e) The final kinetic energy of the alpha particle is, 1. Pas Kasama (-1.40x10-19) 2(4(1.67x10-2) = 1.467x10-12 J =1.47x10-12 J The final kinetic energy of gold nucles is, PA KANJ 27 (2.86x10-19) 2((197)(1.67x10-27)) = 1.24x10-13 J Using conservation of energy is,

kq9 = K. 9x10°)(2)(1.6x10-1)(79)(1.6x10-1) 1.58598x10-12 = 2.295 x 10-44 m = 2.30x10 m The energy pricipal (1) The radius is, l'a+rau = 2x10-15 +8x10-15 = 1x10-14 m So they will not touch The kinetic energy of the alpha particle is, evlev = 1.58598x10-12 =1.6x10-12 J The initial velocity of the alpha particle is, Ko=(9.9x10Ⓡ ev 1.602x10-19 2(1.59598x10-12) 2m, +2m 2(1.59598x10-12) 12(1.67x10-2) + 2(1.67x10-2) = 2.18595x10 m/s =2.2 m/s

(k) The time is, tad 1x 10-13 2.18595x10 = 4.57 x 10-21 s