Question

A 14C nucleus, initially at rest, emits a beta particle. The beta particle is an electron with 133 keV of kinetic energy.

Part A What is the speed of the beta particle? v v = nothing   c   SubmitRequest Answer Part B What is the momentum of the beta particle? p p = nothing   kg⋅m/s   SubmitRequest Answer Part C What is the momentum of the nucleus after it emits the beta particle? p p = nothing   kg⋅m/s   SubmitRequest Answer Part D What is the speed of the nucleus after it emits the beta particle? v v = nothing   m/s

Answer 1

PART A: Rest mass energy of relativistic particle is given by E. = moc? Here to = rest ass, ca speed of light in free space We know, rest mass of electron mo = 9.1x10 3 1 kg and c= 3x108 mis E = (9.1x1639 kg) (3x108m 15 )2 = 8:19 x10'4J Given, Kinetic energy k = 133 Kev = 133x10" ev = (1.33x10 (1-6x109) J = 12-128 x16'4 J Total energy E = K+E. = 8.19 x10h45 +2.128x1014 = 10.318x164 g But E=mc² m=E = 10.31 10.318x10 .10.318x1014, s. " 2 (3 x 108 m/s) = 9x106 mils? We know m=mo - - w - mot v = c fi- moto 14 / 15 x 10 30 kg) Here v= Speed of particle v = (1918 wt) 1 - foto - 135 kvo mys] PART B: Momentum p= mv = (1.15 x 1630 kg) (1.835x108 m/s) 12. ARTC: Conservation of Momentum Momentum of nucleus 1. before emission of ßparticle = Momentum of + Momentum nucleus after of ß-particle 1. emission 0 - +2.11 x 1022 Momentum of nucleus after emission Momentum of nucleus after emission = -2.11 X 1022 kgm/s

PART . rest mass of 14C nucleus -16% rest mass of profon) +(8x rest mass of heutron) +(6x1.67 X1627kg) + (8x 1.67X1079) ~ 2.34 x 1526 kg Momentum b = mv mo 19 b >> F biela movies =w² IM + m2 alus 1 speed of 14C nucleus v=-(2.11x10 2² kg m/s) [1:2-11x10 kquals)" + (2.34x1026 kg) 2 (3x108 N =-0.902 x 104 - 9.02 X10% ms]