44.46: In Eq.(44.9), Ea (m 0 mK 0 )c 2 , and with M
K Ea2 (m c 2 )2 2mp c
2
mp , m
m and Em
(m )c 2
K,
(mp c 2 ) 2
( m )c 2 (938.3 MeV) 2 139.6MeV
(1193 MeV 497.7 MeV) 2 (139.6 MeV) 2 2(938.3 MeV) 904 MeV.
44.55: a) E
( m)c 2
(m )c 2
( m 0 )c 2
(m )c 2
1321MeV 1116 MeV 139.6 MeV E 65 MeV. b) Using (nonrelativistic) conservation of momentum and energy: P 0
m 0v
0
0
Pf
m
0
v
v
m m
0
v 0.
Also K
K
0
E from part (a).
1 m v2 2 K
0
So K
1
44.52: a) The baryon number is 0, the charge is e , the strangeness is 1, all lepton numbers are zero, and the particle is K . b) The baryon number is 0, the charge is e , the strangeness is 0, all lepton numbers are zero, and the particle is . c) Th
v0 vcm . 1 v0 vcm c 2 vcm , v 0, so v M vcm . b) The condition for no net momentum in the For mass M , u center of mass frame is mm vm MM vM 0, where m and M correspond to the velocities )0 M , where found in part (a). The algebra reduces to m m ( 0
44.57: From Pr.(44.56): r R R So
r
.
dR 1 dr r d 1 dr d 2 since 0. dt dt dt dt dt 1 dR 1 dr 1 dr dr 1 dR v So r H 0 r. R dt R dt r dt dt R dt dv d r dR d dR 0 Now d d R dt d dt dR K where K is a constant. dt dR K