TEST BANK FOR Introduction to Elementary Particles 2nd Edition By Griffiths D.J

1 Historical Introduction to the Elementary Particles 1
2 Elementary Particle Dynamics 9
3 Relativistic Kinematics 17
4 Symmetries 37
5 Bound States 57
6 The Feynman Calculus 79
7 Quantum Electrodynamics 97
8 Electrodynamics and Chromodynamics of Quarks 147
9 Weak Interactions 171
10 Gauge Theories 209
Contents VII
11 Neutrino Oscillations 233
12 What’s Next 237
A The Dirac Delta Function 247
1
1
Historical Introduction to the Elementary Particles
Problem 1.1
For an undeflected charged particle, qE = qvB =) v = E
B
.
With just a magnetic field, qvB = m
v2
R
=)
q
m
= v
BR
= E
B2R
.
Problem 1.2
r0 = 10−15 m; ¯h = 6.58 × 10−22 MeV s; c = 3.00 × 108 m/s;
so m = ¯h
2r0c
=

¯hc
2r0

1
c2 = 98.7 MeV/c2 .
Observed mp = 138 MeV/c2. Off by a factor of 1.4.
Problem 1.3
r0 = 10−15 m; ¯h = 6.58 × 10−22 MeV s; c = 3.00 × 108 m/s;
me = 0.511 MeV/c2.
DxDp
¯h
2
so pmin = ¯h
2r0
=

¯hc
2r0

1
c
= 98.7 MeV/c .
Emin =
q
p2
minc2 + m2e
c4 = 98.7 MeV .
The energy of an electron emitted in the beta decay of tritium is < 17 keV.
2 1 Historical Introduction to the Elementary Particles
Problem 1.4
mL = 1
3
[2 (mN + mX) − mS] .
mN = 938.9; mX = 1318.1; mS = 1190.5.
So mL = 1
3
[2 (2257.0) − 1190.5] = 1107.8 MeV/c2.
Observed mL = 1115.7 MeV/c2. Off by 0.7%.
Problem 1.5
m2
h = 1
3
h
2

m2
K + m2
K

− m2
p
i
= 1
3

4m2
K − m2
p

.
mK = 495.67; mp = 138.04.
m2
h = 1
3
h
9.637 × 105
i
= 3.212 × 105 ) mh = 566.8 MeV/c2.
Actually mh = 547.3 MeV/c2. Off by 3.5%.
Problem 1.6
MD − MS = 1232 − 1385 = −153
MS − MX = 1385 − 1533 = −148. Average: −151.
) MW = MX + 151 = 1533 + 151 = 1684 MeV/c2.
Actually MW = 1672 MeV/c2. Off by 0.7%.
Problem 1.7
(a) D− −! n + p− or S− + K0
S+ −! p + K¯0; S+ + p0; S+ + h; S0 + p+; L + p+; X0 + K+
X− −! S0 + K−; S− + K¯0; L + K−; X0 + p−; X− + p0; X− + h
3
(b) Kinematically allowed:
D− −! n + p−
S+ −! S+ + p0; S0 + p+; L + p+
X− −! X0 + p−; X− + p0
Problem 1.8
(a) With a strangeness of −3, the W− would have to go to
􀀀
X0 + K−
or
X− + K¯0

to conserve S and Q. But the XK combination is too heavy
(at least 1808 MeV/c2, whereas the W− is predicted – see Problem 1.6 –
to have a mass of only 1684).
(b) About 0.5 cm ; t = d/v = (5 × 10−3 m)/(3 × 107 m/s) = 2 × 10−10 s.
(Actually, t = 0.8 × 10−10 s.)
Problem 1.9
S+ − S− = 1189.4 − 1197.4 = −8.0
p − n + X0 − X− = 938.3 − 939.6 + 1314.8 − 1321.3 = −7.8

3% difference.
Problem 1.10
Roos lists a total of 30 meson types; in the first column is the particle name at
the time, in the second column the quoted mass (in MeV/c2), and in the third
its current status.
4