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

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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

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