Appendix Solutions of Exercises and Problems

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A.1 Solutions of Nuclear Physics (Chapter 1)

1.1 Initial Problems

Exercise 1.1.1 To give a rough estimate of the nuclear density, we assume that – the binding energy is negligible; – proton and neutron have the same mass, m p = mn; 1/3 – the nuclear radius is R = r0 · A , with r0 = 1.2 fm. Under these assumptions we have

M A · m 3 m 3 × 1.67 10−24g ρ = = p = p 2.3 · 1014 g/cm3. / π 3 π 3 . ( . −13 )3 V 4 3 r0 A 4 r0 12 56 1 210 cm

Exercise 1.1.2 The electrostatic energy for a charge Q distributed uniformly in a sphere of radius 2 R is 3/5 · Q /(4π0 R). Equating this energy to the Coulomb binding energy in the SEMF we get , 3 Z 2e2 Z 2 = a · , 1/3 C 1/3 20 π0 r0 A A and then

3 e2 1 3 1 197 MeV fm 1 aC = × × = × αc × 0.6 × × 0.7MeV. 5 4π0 r0 5 r0 137 1.2fm

© Springer Nature Switzerland AG 2019 43 S. Petrera, Problems and Solutions in Nuclear and Particle Physics, UNITEXT for Physics, https://doi.org/10.1007/978-3-030-19773-5 44 Appendix: Solutions of Exercises and Problems

Exercise 1.1.3 Using the result of problem 1.1.1, we assume for the nuclear density ρ 2.3 × 1014 g/cm3. Denoting with R and M respectively the radius and mass of the neutron star, from the relation 4 3 π R ρ = M ≈ M 3 we obtain 1/3 33 1/3 3M 3 · 210 g R ≈ 12.8km. 4πρ 4 · 3.14 · 2.31014 g/cm3

Exercise 1.1.4 Let us consider two deuterons moving along a certain direction with equal but opposite velocities (head-on collision). Since the motion is thermal, the kinetic energy of each deuteron can be treated as non-relativistic, E = 1/2Mv2, and assumed to be of the order of kB T . At large distance, in the rest frame of one of the deuterons, the other has velocity 2v. The corresponding kinetic energy equates the repulsive electrostatic energy at the minimum distance, because of energy conservation 2 1 2 1 e M(2v) = 4E = 4kB T = . 2 4π0 rmin

Thus a rough estimate of the minimum temperature to get nuclear processes is e2 1 αc Tmin = = 4π0 4kBrmin 4kBrmin 197 MeV · fm 4109 K. 137 × 4 × 8.610−11MeV · K−1 × 1fm

Exercise 1.1.5 The neutron rate per solid angle is

dN dσ dn = b n L dtd d dt T where dnb/dt = I/e is the deuteron beam intensity and nT is the number of target nuclei per unit volume, nT = NA/A ρ. The solid angle between the detector and the interaction region (assumed point-like) is = S/R2. Then we have

dN dσ S I N = A ρL dt d R2 e A

20 210−6 A 61023 13 10−3 10−24cm2/sr 0.210−3 g/cm2 1.4103 s−1 3002 1.610−19C 3 Appendix: Solutions of Exercises and Problems 45

1.2 Nuclear Scattering

Exercise 1.2.1 (1) The rate of electrons scattered in the solid angle around the angle θ, from a beam of intensity dnb/dt (e/s) incident perpendicularly on a target with atomic 2 number A, thick xT (g/cm ),is dn dnb dnT dσ Ie NA S dσ = × × d × xT × × (θ), dt dt dS d e A R2 d √ being S/R 1. Then we have

−6 . · . 23 σ dn 510 × 0 12 6 02 10 × 20 × d (θ) dt 1.610−19 40 1002 d sr dσ 1.13 1032 × (θ). cm2 · s d

2 2 dσ/d(θ) is given by |F(q )| × (dσ/d)Mott.Forβ → 1 the Mott cross section ◦ at 40 is σ 2α2( )2 2 θ/ × 2 d = Z c cos 2 20 197 ( )2 4 θ/ d Mott 4 pc sin 2 137

cos2 20◦ × fm2/sr 0.272 mb/sr 4 × 7002 × sin4 20◦

The form factor for a uniform charge distribution in a sphere of radius RA is

sin x − x cos x F(q2) = 3 , x3 where x = qRA/

2 pc sin θ/2 × (1.18 A1/3 − 0.48) fm x = c

2 × 700 MeV × sin 20◦ × 3.56 fm 8.64, 197 MeV fm

Hence we get F(q2) 3.18 10−2 and ﬁnally obtain dn sr cm2 1.13 1032 × 0.272 10−27 × (3.18 10−2)2 31 electrons/s. dt cm2 · s sr In Fig. 1.1 the rate of the scattered electrons is shown as a function of the angle. 46 Appendix: Solutions of Exercises and Problems

Fig. 1.1 Rate (counts-per-sec) for 700 MeV/c electron scattering 106 against 40C

105

104

103

102

10 15 20 25 30 35 40 45 50 θ [degrees]

(2) As it can be seen in the ﬁgure the ﬁrst local maximum is at about 25◦. Here the detector delivers about 1400 counts per second. The mean number of electron-ion pairs produced by an electron crossing the gas mixture is

(−dE/dx)ion × ρ × d Ne = × ion Wion

1.4 × 2106 eV/(gcm−2) × 1.810−3g/cm3 × 0.1cm × 0.10 3.36, 15 eV whereweused2MeV/(gcm−2) for the minimum ionization energy loss. The number of events for which no electron reaches the anode is

Ne 3.36 0 = (1 − P) 0.70 30.2%.

The rate of coincident counts is ﬁnally

dn dn c = × (1 − )2 1400 × 0.6982 1400 × 0.49 690 counts/s. dt dt 0

Exercise 1.2.2 The number of minima is given by number of the zeroes of the form factor for a uniform charge distribution. The latter is given by

sin x − x cos x F(q2) = 3 , x3 Appendix: Solutions of Exercises and Problems 47

Fig. 1.2 tan x versus x 12 (black). y = x (red) 10

0 π 3π 5π 7π 9π 2 2 2 2 2 −2

0246810121416

where x = qR/, with R given by the radius of the nucleus, R = (1.18A1/3 − 0.48) fm. F(q2) = 0 leads to the equation

tan x = x.

A graphical method allows to estimate the positions of the zeroes (see Fig. 1.2, black: tan x, red: x) as the ones where the tangent equates the straight line. This occurs close to x 3π/2, 5π/2, 7π/2, 9π/2 ... In the actual experimental conditions x is limited up to a maximum xmax = qmax R/. Remembering that

θ E q = 2 p · sin =⇒ q = 2 p 2 2 max c we have E 2 · 180 MeV x = 2 × (1.18 A1/3 − 0.48) fm × 6.4fm 11.7. max c 197 MeV fm There are three minima below this value, corresponding to the zeroes up to 7π/2.

Exercise 1.2.3 Considering the Rutherford cross section, we can write the counting rate at angle θ as

f (θ) = K , sin4 θ/2 48 Appendix: Solutions of Exercises and Problems where is the incident ﬂux and K an overall factor including various terms (kinematical, geometrical, etc.). We assume that

f (20◦) = K = 1s−1. (1.1) sin4(20◦/2)

Denoting with fa the counting rate for a ﬂux attenuated by a factor a,wehave

◦ a −1 fa(10 ) = K = 1s , (1.2) sin4(10◦/2)

Dividing (1.2)by(1.1) we get sin 5◦ 4 a = 6.3%. sin 10◦

◦ ◦ Using the attenuated beam, the counting rate at 20 is fa(20 ). The mean waiting time is its inverse 1 1 1 t = = = 16 s. ◦ ◦ −1 fa(20 ) af(20 ) 0.063 × 1s

Exercise 1.2.4 2 2 The differential cross section dσ/d(θ) is given by |F(q )| × (dσ/d)Mott.For β → 1 the latter cross section is σ 2α2( )2 2 θ/ d = Z c cos 2 ( )2 4 θ/ d Mott 4 pc sin 2 6 × 197 2 cos2 7.5◦ × fm2/sr 6.3 × 10−26 cm2/sr. 137 4 × 1002 × sin4 7.5◦

√The form factor can be neglected because the momentum transfer is small.1 Since S/R 1, we can simply write dσ σ = × S/R2 . · −26 2/ × . · −4 µ . 6 3 10 cm sr 7 5 10 sr 47 b d Mott

1 1/3 The relevant argument for the form factor is x = qRA/,whereRA 1.2fm× A is the nuclear radius. Hence x 0.36 and for a uniform charge distribution, we have sin x − x cos x F(q2) = 3 0.99. x3 . Appendix: Solutions of Exercises and Problems 49

22 −2 The number of scatterers per unit surface is dnT/dS = d × NA/A 5 × 10 cm . Then we have dN I dn e = 0 σ T 1.5 × 108 s−1. dt e dS Exercise 1.2.5 The momentum transfer (we have pc E)is

q = 2 p · sin θ/2 = 2 × 500 × sin 5◦ 87.2MeV/c.

The Mott cross section at 10◦ can be written as dσ Z 2α2( c)2 θ = 4 E2 cos2 ( )4 d Mott qc 2 26 × 197 2 5002 cos2 5◦ 4 × × fm2/sr 0.24 b/sr 137 87.24

The form factor is given by sin x − x cos x F(q2) = 3 , x3

1/3 where x = qRA/.UsingRA 1.2fm×A for the nucleus radius, we ﬁnd x = 87.2 × 4.6 / 197 2 and then dσ dσ b b = |F(q2)|2 = . × . 2 . 0 24 0 65 0 10 d d Mott sr sr

Exercise 1.2.6 The Rutherford cross section can be written as σ α( ) 2 d = zZ c 1 , θ d 4Eα 4 sin 2 where z and Eα are respectively the charge and kinetic energy of alpha particles. The solid angle corresponding to the detector is

S = = 10−3 sr. R2 To achieve the required accuracy, we calculate the cross section at the largest angle (150◦) in the chosen interval, where it has the smallest value α( ) 2 × × 2 −3 σ = zZ c 2 79 197 10 . 2 = . × −27 2. θ 0 12 fm 1 2 10 cm 4Eα 4 137 × 4 × 5.5 0.8705 sin 2 50 Appendix: Solutions of Exercises and Problems

The event rate in the detector is

NA r = Iαρl σ A

23 −1 where Iα is the beam intensity (NA = 6.02 × 10 mole is the Avogadro number). Thus the intensity of the α beam must be

A r 197 10 7 −1 Iα > = 2.7 × 10 s . 23 −27 2 ρlNA σ 0.1 × 6.02 × 10 1.2 × 10 cm

Exercise 1.2.7 + 7 → 4 4 The Q-factor of the reaction p 3Li 2He + 2He is

= + − = +[ + − (7 )]− [ + − (α)]= Q Mp MLi 2Mα Mp 3Mp 4Mn B 3Li 2 2Mp 2Mn B

= (α) − (7 ) = · . − . = . > 2B B 3Li 2 28 3 39 3 17 3MeV 0

The reaction is exothermic. 7 According to the shell model, the 3Li shell occupancies for protons and neutrons are p : (1s1/2)2(1p3/2)1

n : (1s1/2)2(1p3/2)2

The spin-parity is then determined by the odd (1p3/2) proton shell and is J P = (3/2)−. 7 Protons at rest cannot interact with 3Li nuclei because of the Coulomb barrier. 7 Neglecting for simplicity the size of the proton with respect to the one of 3Li, the minimum proton kinetic energy turns out to be

zZe2 zZαc 3 · 197 MeV fm T = 1.9MeV. min π (7 ) · . 1/3 4 0d R 3Li 137 1 27 fm

Indicating the spin-parities of the nuclei involved in the reaction, we have 1 + 3 − p + 7Li → 4He(0+) + 4He(0+) 2 3 2 2 2

Knowing that the ﬁnal orbital angular momentum is zero, we deduce that the initial total angular momentum must be zero. The angular momentum conservation imposes

1 3 ⊕ ⊕ L = 0 2 2 i Appendix: Solutions of Exercises and Problems 51 denoting with ⊕ the operation of addition of angular momenta and with Li the initial 1 ⊕ 3 = , orbital angular momentum. Since 2 2 1 2, then it follows that Li must be either 1or2. On the other hand parity conservation imposes the same parity for the initial and ﬁnal states. The ﬁnal parity is evidently +1 and then

= ( ) × (7 ) × = (+ ) × (− ) × (− )Li Pi P p P 3Li Porb 1 1 1 hence Li must be odd and ﬁnally Li = 1. Exercise 1.2.8 The Mott cross section at 5◦ is σ α 2 2 θ/ d = Z c cos 2 4 θ/ d Mott pc 4sin 2 6 × 197 2 cos2 2.5◦ fm2 mb × 99 137 × 720 4 × sin4 2.5◦ sr sr

Using the measured cross section, we derive the absolute square of the form factor as dσ 80 |F(q2)|2 = d meas 0.808 dσ 99 d Mott

If the nucleus is spherically symmetric, the form factor is real and then it can be obtained as the square root of its absolute square. Note that the indetermination of the sign is resolved looking at the momentum transfer. At 5◦ it is small and the form factor is still far from the ﬁrst zero. Thus we can assume that the form factor is positive. The momentum transfer turns out to be θ q = 2 p sin 62.8MeV/c 2 and then

62 6 1972 √ r 2 = [1 − F(q2)] × (1 − 0.808) 5.95 fm2 q2 62.82 from which we get for the 12C nuclear radius 2.44 fm. Exercise 1.2.9 For β 0.1 a proton is non-relativistic and its maximum energy can be written as

1 T m β2 0.5 × 938 × 0.01 4.7MeV. (1.3) max 2 p 52 Appendix: Solutions of Exercises and Problems

The maximum value of beta (and then of energy) corresponds to the forward scattering with the incident particle (γ or n) scattered back. In the case of photon scattering, γ + p → γ + p, the kinematic relation is the same as in the Compton scattering, with the proton replacing the electron.

Eγ Eγ = , 1 + Eγ /m p(1 − cos θ)

where Eγ and Eγ are the photon initial and ﬁnal energies and θ is the photon scattering ◦ angle. For θ = 180 the proton energy is maximum. Denoting with T = Eγ − Eγ the proton kinetic energy, its maximum is

2 Eγ 2Eγ Tmax = Eγ − = (1.4) 1 + 2Eγ /m p m p + 2Eγ

(a) If we assume 10 MeV for the photon energy, from (1.4)wehave

2 2Eγ 2 × 100 Tmax = 0.21 MeV, m p + 2Eγ 938 + 20 which is largely smaller than (1.3). (b) Solving Eq. (1.4)inEγ and assuming that the proton energy is the one measured (1.3), the photon energy is √ Tmax + Tmax(Tmax + 2m p) 4.7 + 4.7(4.7 + 2 · 938) Eγ = 49 MeV. 2 2 (c) In the case of neutron scattering, n + p → n + p, the maximum proton energy occurs again in the forward scattering. Since the proton and neutron masses are very similar, this corresponds to the neutron stopped and the proton achieving the entire initial energy. This is a well known fact in a billiard, but can be easily obtained from kinematics. Being θ = 0, we can treat the kinematics in one dimension as

= + = + . Tn Tn Tp pn pn pp

2 Expressing the kinetic energies as T = p /(2m) and assuming mn = m p, we get

( − ) = =⇒ = . . pp pp pn 0 pp pn 4 7MeV

Exercise 1.2.10 The Rutherford cross section is σ α( ) 2 d = zZ c 1 , θ d 4Eα 4 sin 2 Appendix: Solutions of Exercises and Problems 53 where z and Eα are respectively the charge and kinetic energy of alpha particles. Using the solid angle subtended by the detector

S 0.5 = = = 510−3 sr, R2 102 we get for the cross section at angle θ α( ) 2 × × 2 −3 σ(θ) = zZ c 2 79 197 510 θ θ 4Eα 4 137 × 4 × 5.64 4 sin 2 sin 2

0.507 5.07 × 10−27 2 = 2. 4 θ fm 4 θ cm sin 2 sin 2

The number of counts during the time interval t is

NA N(θ) = Iαtρd σ(θ) A where Iα is the α beam intensity and d is the target thickness. Solving in Iα we have

(θ) θ = N = 197 × (θ) 4 Iα − N sin ρ NA σ(θ) 0.005 × 19.3 × 3600 × 6.02 1023 × 5.07 10 27 2 d t A and then θ 4 −1 Iα N(θ) × 186 × sin s . 2

The following table gives the beam intensity in s−1 resulting at each angle

θ 15◦ 25◦ 35◦ 45◦ 55◦ 65◦ 75◦ − Iα (s 1) 229.9 ± 3.5 242.1 ± 9.9 226 ± 19 199 ± 28 262 ± 47 201 ± 56 179 ± 67

√ The statistical errors are calculated as N(θ) × 186 × sin4 θ/2. Calculating the weighted average and its variance we get

−1 Iα = 230.7 ± 3.2s . 54 Appendix: Solutions of Exercises and Problems

1.3 Nuclear Binding Energy

Exercise 1.3.1 Decays among isobar nuclei belong to the class of beta decays. In the present case the mass number, A = 197, is odd so that there is only one stable nucleus. In fact, using the semi-empirical mass formula (SEMF) the atomic mass M(A, Z) as a function of Z is a single curve, because the pairing term is null for all isobars. The stable − nucleus has Zs = 79, the nucleus with Z = Zs − 1 = 78 can transmute to it via β + decay whereas the nucleus with Z = Zs + 1 = 80 can do it via β decay or electron capture (EC). We can write the atomic mass of A = 197 nuclei as

2 M(197, Z) = Zmp + (197 − Z)mn − B(197, Z)/c + Zme, where B(197, Z) is the nuclear binding energy, for which we use the SEMF. Writing explicitly only the terms depending on Z we have

2 ( − )2 2 2 Z 197 2Z M(197, Z) c = const + Z (m p − mn + me) c + aC + aA 1971/3 197

Z 2 (197 − 2Z)2 const − 0.782 Z + 0.697 + 23.3 MeV. 5.82 197

β− 197 M( , ) − M( , ) . For the transition from 78 Pt we have 197 78 197 79 0 90 MeV, hence it is allowed. β+ 197 M( , ) − M( , )> , The transition from 80 Hg is allowed if 197 80 197 79 2me if instead this difference is positive only the electron capture is possible. In the present case we have M(197, 80) − M(197, 79) 0.3 MeV/c2. As a conclusion the possible decay types are

β− : 197 → 197 + − +¯ν 78 Pt 79 Au e e

: − + 197 → 197 + ν EC e 80 Hg 79 Au e

Exercise 1.3.2

The mean neutron kinetic energy is E ≈kB T kB 300 25 meV. From the semi-empirical mass formula we get

B(235, 92) = 1786.8MeV

B(148, 57) = 1209.8MeV

B(87, 35) = 745.4MeV. Appendix: Solutions of Exercises and Problems 55

The neutron energy is negligible with respect to the other energies and then we have for the energy release

Q = B(148, 57) + B(87, 35) − B(235, 92) 168 MeV.

A similar though less accurate conclusion can be reached using the B/A values from the binding energy per nucleon plot reported in all the textbooks. The values are 7.6, 8.2 and 8.6 MeV, respectively A = 235, 148 e 87. Hence we obtain

Q = 148 · 8.2 + 87 · 8.6 − 235 · 7.6 176 MeV.

Exercise 1.3.3 At large distance, in the rest frame of one of the nuclei, the other has velocity 2v.At the minimum distance R the two nuclei are at rest. From energy conservation then we get 1 e2 αc 197 MeV fm M(2v)2 = 4E = = = 1MeV, 2 4π0 R R 137 · 1.4fm

E, the mean kinetic energy of each nucleus, is then about 0.25 MeV. Knowing that for T = 300 K the mean kinetic energy is kB T 25 meV, the temperature for which nuclei have E 0.25 MeV is

E 0.25 106 eV × 300 T = = 3109 K. −3 kB 25 10 eV

The energy release is Q = BT − 2BD 4MeV.

Exercise 1.3.4 The reaction in the text belongs to the more general class

− νe + (A, Z) → (A, Z + 1) + e .

The threshold energy is given by

(m + M )2 − M2 E = e , (1.5) th 2M where M and M are the masses of (A, Z) and (A, Z + 1) nuclei respectively. These masses are related to the binding energies as follows

2 M = ZMp + (A − Z)Mn − B(A, Z)/c

2 M = (Z + 1)Mp + (A − Z − 1)Mn − B(A, Z + 1)/c = M + M, 56 Appendix: Solutions of Exercises and Problems hence we have

2 M = (Mp − Mn) + B/c with B = B(A, Z) − B(A, Z + 1).

B can be calculated using the semi-empirical mass formula. In particular for odd-A nuclei only the Coulomb and asymmetry terms are needed, because: 1. the volume and surface terms depend only on A and they cancel out in the difference; 2. for odd-A the pairing term is null for both initial and ﬁnal nuclei. We have

Z 2 (Z + 1)2 (A − 2Z)2 [A − 2(Z + 1)]2 B =−a − − a − = C A1/3 A1/3 A A A

2Z + 1 A − 2Z − 1 = a − 4a C A1/3 A A In the reaction considered in the text A = 37 and Z = 17 and then we have 35 2 M =−1.293 + 0.697 × − 4 × 23.3 × 1MeV. 371/3 37 Substituting this value in (1.5) we obtain

[m + (M + M)]2 − M2 m (m + 2M) + M(2m + M + 2M) E = e = e e e th 2M 2M

me + M 1.5MeV.

Exercise 1.3.5

Denoting by Q− the Q-factor for the β− decay

64 → 64 + − +¯ν 29Cu 30Zn e e and Q+ the one for the β+ decay

64 → 64 + + + ν 29Cu 28Ni e e we have (omitting the factor c2 in the mass terms)

Q− = 29Mp + 35Mn − B(64, 29) − 30Mp − 34Mn + B(64, 30) − me

= Mn − Mp − me + B(64, 30) − B(64, 29) 0.782 MeV + B(64, 30) − B(64, 29). Appendix: Solutions of Exercises and Problems 57

Similarlywehave

Q+ = Mp − Mn − me + B(64, 28) − B(64, 29) −1.804 MeV + B(64, 28) − B(64, 29).

From the SEMF we get

302 − 292 (64 − 60)2 − (64 − 58)2 B(64, 30) − B(64, 29) =−0.697 × − 23.3 × + 641/3 64

12 + 12 + √ 0.0005 MeV, 64

282 − 292 (64 − 56)2 − (64 − 58)2 B(64, 28) − B(64, 29) =−0.697 × − 23.3 × + 641/3 64 12 + 12 + √ 2.74 MeV. 64

Hence we have Q− 0.78 MeV Q+ 0.94 MeV.

Both decays are allowed. The maximum kinetic energies of the electron and positron are equal respectively to Q− and Q+. Exercise 1.3.6 ∂M(A,Z) = M( , ) The stability condition can be written as ∂ Z 0, A Z being the atomic mass of the nucleus (A, Z), which is is single function for odd-A nuclei. Using the SEMF we have 2a Z 4a (A − 2Z) c − a − (M − M − m )c2 = 0 A1/3 A n p e hence we get for the asymmetry coefﬁcient A 2a Z a = c − (M − M − m )c2 24 MeV. A 4(A − 2Z) A1/3 n p e

Exercise 1.3.7 + + For a β decay, (A, Z) → (A, Z − 1) + e + νe,theQβ value, is

2 Qβ =[M(A, Z) − M(A, Z − 1) − m]c , where 2 M(A, Z) = ZMp + (A − Z)Mn − B(A, Z)/c

2 M(A, Z − 1) = (Z − 1)Mp + (A − Z + 1)Mn − B(A, Z − 1)/c . 58 Appendix: Solutions of Exercises and Problems

Hence we have 2 Qβ =[Mp − Mn − m]c − B, (1.6) where B = B(A, Z) − B(A, Z − 1).

Calculating B from the SEMF, we observe that all terms cancel but the Coulomb and asymmetry ones because 1. the volume and surface terms depend on A only, 2. A is odd and the pairing term is the same (=0) for both nuclei. Then we have

Z 2 (Z − 1)2 (A − 2Z)2 [A − 2(Z − 1)]2 B =−a − − a − C A1/3 A1/3 A A A

2Z − 1 A − 2Z + 1 =−a − 4a (1.7) C A1/3 A A Considering the decay in the text, we have A = 35 and Z = 18: hence the term multiplying aA vanishes. Inverting equation (1.7) we obtain

A1/3B a =− . C 2Z − 1

From (1.6)wehave

2 B =[Mp − Mn − m]c − Qβ =−1.293 − 0.511 − 4.95 −6.75 MeV, where we have used the maximum positron energy for Qβ . Finally we get

351/3 ×−6.75 a =− 0.63 MeV. C 35 The value obtained in this way differs from the best-ﬁt value (0.697 MeV) given with the SEMF by less than 10%.

Exercise 1.3.8

Denoting by Q− the Q-value for

100 → 100 + − +¯ν 43 Tc 44 Ru e e and with Q+ the one for

100 → 100 + + + ν 43 Tc 42 Mo e e Appendix: Solutions of Exercises and Problems 59 we have (omitting c2 multiplying the masses)

Q− = 43Mp + 57Mn − B(100, 43) − 44Mp − 56Mn + B(100, 44) − me =

= Mn − Mp − me + B(100, 44) − B(100, 43)

B(100, 44) − B(100, 43) + 0.782 MeV and Q+ = Mp − Mn − me + B(100, 42) − B(100, 43)

B(100, 42) − B(100, 43) − 1.804 MeV

From the semi-empirical mass formula we have for an odd-A odd-Z nucleus

Z 2 − (Z ± 1)2 B(A, Z) − B(A, Z ± 1) =−a − C A1/3

(A − 2Z)2 −[A − 2(Z ± 1)]2 a −a − 2 P . A A A1/2 In the current case we have

432 − 442 B(100, 43) − B(100, 44) =−0.697 × − 1001/3

(100 − 86)2 − (100 − 88)2 12 −23.3 × − 2 × √ −1.45 MeV 100 100

432 − 422 B(100, 43) − B(100, 42) =−0.697 × − 1001/3

(100 − 86)2 − (100 − 84)2 12 −23.3 × − 2 × √ −1.18 MeV. 100 100

Hence we obtain Q− 2.23 MeV Q+ −0.62 MeV.

β− 100 β+ The decay to 44 Ru is allowed. Instead the decay is forbidden, yet the electron 100 = + capture, allowing the transition to 42 Mo, is possible since we have Q EC Q+ 2me 0.40 MeV > 0.

Exercise 1.3.9 (a) Each ﬁssion reaction releases 200 MeV = 2 108 eV × 1.6 10−19 J/eV 3.2 10−11 J. Hence the ﬁssion rate is 60 Appendix: Solutions of Exercises and Problems

P 2109 r = = 6.25 1019 s−1. −11 Eﬁss 3.210

(b) 1 g of 235U releases an energy

N 61023 E × A = 3.210−11 × 0.81011 J/g. ﬁss A 235 In a year the total energy is

J 2109 × 3.15 107 s 6.31016 J. s

The 235U mass consumed in a year is then

6.31016 J M(235U) = 7.88 105 g = 788 kg. 0.81011 J/g

Since 235U is about 30%, the used fuel mass is about 2.6 ton. (c) The maximum neutrino energy is equal to the Q-factor of the beta decay. Denoting 145 β− 2 it by Q−,forthe57 La decay, we have (omitting c in the mass terms):

Q− = Mn − Mp − me − B(145, 57) + B(145, 58) 0.782 MeV − B− where B− is the difference in binding energy between parent and daughter nuclei. Using the SEMF we have for odd-A nuclei

B− = B(A, Z) − B(A, Z + 1)

Z 2 − (Z + 1)2 (A − 2Z)2 −[A − 2(Z + 1)]2 −a − a , C A1/3 A A which becomes in our case

572 − 582 (145 − 114)2 − (145 − 116)2 B− =−0.697 × − 23.3 × −4.03 MeV 1451/3 145 The maximum neutrino energy is then 0.782 + 4.03 4.81 MeV. (d) The neutrino intensity is 20% of the ﬁssion rate 0.20 × 6.25 1019 s−1 = 1.25 1019 s−1. At 500m distance the neutrino ﬂux is

19 Iν 1.25 10 = 4 × 1012 m−2s−1 4π R2 12.56 5002 (e) For a detector having a length l (along the neutrino direction), a section S, com- posed of material of atomic mass A, the interaction rate is Appendix: Solutions of Exercises and Problems 61

N N r = × σ × A × ρlS = × σ × A × M. A A Such proportionality between rate and mass is holding each time the detector length is much smaller of the interaction length. Inserting our values we get

61023 1.44 10−5 450 r = 4 × 1012 m−2s−1 × 610−48m2 × × 106 g s−1 yr−1 gA A A

Exercise 1.3.10 β+ 27 2 Let us call Q+ the Q-factor for the -decay of 14Si. Omitting the factor c multiplying the mass terms, we have:

(Ee)max = Q+ = Mp − Mn − me − B(27, 14) + B(27, 13)

27 (a) The binding energy of 13Al is then

B(27, 14) = B(27, 13) + Mp − Mn − me − (Ee)max = 219.36 MeV

(b) The nuclei involved in the decay are odd-A, hence the pairing term of the SEMF disappears from the mass difference. The volume and surface terms do not contribute in any case. Considering the two surviving terms (aC and aA), the asymmetry term does not contribute since we have, for A = 27 and Z = 14, (A − 2Z)2 = (A − 2 2(Z − 1)) . Hence the mass difference depends only on the Coulomb term aC . (c) For a uniform charge distribution, we have

3 e2 3 αc B = [Z 2(Si) − Z 2(Al)]= [Z 2(Si) − Z 2(Al)]. 5 4π0 R 5 R

27 Hence we get for the 14Si radius

3 αc 1.4MeVfm R = [Z 2(Si) − Z 2(Al)]=0.6 × × (142 − 132) 4.1fm 5 B 5.59 MeV Exercise 1.3.11 ν 71 → 71 − The reaction e + 31Ga 32Ge + e belongs to the more general class

− νe + (A, Z) → (A, Z + 1) + e , for which the neutrino threshold energy is

(m + M )2 − M2 E = e , th 2M 62 Appendix: Solutions of Exercises and Problems where M and M are the masses of the nuclei (A, Z) and (A, Z + 1) respectively. Denoting by M the mass difference M − M,wehave

[m + (M + M)]2 − M2 E = e th 2M which, being M, me M, becomes

Eth me + M. (1.8)

We can write M as

2 M = M − M = Mp − Mn + B/c (1.9) with B = B(A, Z) − B(A, Z + 1).

Combining (1.8) and (1.9) we ﬁnally get

2 B(71, 32) = B(71, 31) − B = B(71, 31) − M − (Mn − Mp)c =

2 = B(71, 31) − Eth − (Mn − Mp − me)c

618.95 − 0.233 − 0.782 617.93 MeV.

Using the SEMF we have instead

B(71, 32) 620.88 MeV which differs about 0.5 percent from the previous value.

Exercise 1.3.12 ∂M(A,Z) = The minimum atomic mass for isobars can be obtained from the equation ∂ Z 0. Using the SEMF we obtain

2a Z 4a (A − 2Z) C − A + (M − M − m )c2 = 0 A1/3 A n p e

The term which depends on the electromagnetic coupling constant is aC . Solving in aC we get A1/3 4a (A − 2Z) a = A − (M − M − m )c2 C 2 Z A n p e

133 If the stable nucleus is 54 Xe, the aC parameter should be Appendix: Solutions of Exercises and Problems 63 1331/3 4 × 23.3 (133 − 108) a = − 0.782 MeV 0.791 MeV. C 2 × 54 133

From classical electrostatics we know that the Coulomb term is proportional to the ﬁne structure constant aC ∝ α. Hence the change in the coupling constant is

α a 0.791 − 0.697 = C 13% α aC 0.697

Exercise 1.3.13 (a) For photons colliding against a ﬁxed iron target, the threshold energy is

( + )2 − 2 μ2 th M m M Eγ = = (1.10) 2M 2M where M is the mass of the initial nucleus, (A, Z), M that of the ﬁnal nucleus, (A − 1, Z), and m the neutron mass. Denoting by M the nuclear mass difference M − M and with B the corresponding binding energy difference, we have

μ2 =[(M − M) + m]2 − M2 = m(m + 2M) − M(2M + 2m − M).

Since M, m M, we get

μ2 2M(m − M) = 2MB, (1.11) having used the relation

M = M − M = m − B

56 55 For 26Fe photo-disintegrating into 26Fe, B can be obtained from the semi-empirical mass formula

B = B(56, 26) − B(55, 26) 490.95 − 478.90 12 MeV. (1.12)

Hence the photon threshold energy is th 2M B Eγ = B 12 MeV. 2M (b) In the case of Cosmic Rays, the collision does not occur in a ﬁxed target frame and the expression (1.10) cannot be used. Instead we make use of the invariance of the total 4-momentum squared so that we can write, at the threshold

( + )2 = ( th + )2 − ( th + )2 = th2 + 2 + th − th 2 − 2 − th · , M m EN Eγ pN pγ EN Eγ 2EN Eγ pN pγ 2 pN pγ 64 Appendix: Solutions of Exercises and Problems

th, th where (EN pN ) is the 4-momentum of the initial nucleus (at the threshold energy) and (Eγ , pγ ) the photon 4-momentum. Since nuclei are ultra-relativistic we have

( + )2 2 + th ( − θ), M m M 2EN Eγ 1 cos where θ is the angle between the nucleus and photon directions. Using μ2 we obtain

th ( − θ) μ2. 2EN Eγ 1 cos

The nucleus threshold energy is

μ2 th = . EN 2Eγ (1 − cos θ)

Using (1.11) and θ = π (head-on collisions), we ﬁnally get

MB 52 GeV 12 MeV th = = × 20 EN − 3 10 eV 2Eγ 210 3 eV where we have used M = M(56, 26) 52 GeV, as obtained from the SEMF.

Exercise 1.3.14 (a) We can write the two separation energies as

Sp = B(A, Z) − B(A − 1, Z − 1)

Sn = B(A, Z) − B(A − 1, Z) hence we obtain for the difference

Sp − Sn = B(A − 1, Z) − B(A − 1, Z − 1).

Using the SEMF we get

Z 2 − (Z − 1)2 (A − 1 − 2Z)2 −[A − 1 − 2(Z − 1)]2 Sp − Sn =−aC − aA + DP = (A − 1)1/3 A − 1

2Z − 1 A − 2Z − a + 4a + D , (1.13) C (A − 1)1/3 A A − 1 P where DP originates from the difference of the pairing terms, δP (A). The possible values of DP are reported in the following table, where e and o stand for even and odd nucleon parity in the corresponding nucleus. (b)IfinEq.(1.13) we insert Z = A/2, valid for light nuclei, we obtain Appendix: Solutions of Exercises and Problems 65

parent Sp : (Z, N − 1) Sn : (Z − 1, N) DP Z, N parity δP parity δP ee eo 0 oe 0 0 oo oe 0 √ eo 0 √ 0 √ eo ee +aP /√A − 1 oo −aP /√A − 1 +2aP /√A − 1 oe oo −aP / A − 1 ee +aP / A − 1 −2aP / A − 1

A − 1 S − S =−a =−a (A − 1)2/3. p n C (A − 1)1/3 C

Note that in this case we have DP = 0, because A = 2Z is necessarily even, corresponding to the ﬁrst two rows of the table. The difference Sp − Sn is always negative and decreasing with A. Hence larger energy is needed to extract neutrons than protons. (c) If instead we use Z = A/2.5, approximately valid for heavy nuclei, we obtain

2/2.5 A − 1 2 A S − S =−a + a + D = p n C (A − 1)1/3 A 2.5 A − 1 P 2 A − 1.25 A = −a + a + D . 2.5 C (A − 1)1/3 A A − 1 P

Also in this case the difference Sp − Sn decreases with A. The general treatment is complicated because of the presence of DP , which can have either sign. For even-A nuclei (DP = 0) the curve starts from positive values because of the aA term. In the following ﬁgure the Sp − Sn behaviour is shown for even-A nuclei.

20 (MeV) n -S p S 10 A = 2.5 Z

−10 A = 2 Z −20

20 40 60 80 100 120 140 160 180 200 220 A 66 Appendix: Solutions of Exercises and Problems

(d) Using (1.13) we get • 20 − − . 10Ne: Sp Sn 4 96 MeV; • 38 − − . 18Ar: Sp Sn 2 28 MeV; • 106 − − . 46 Pd: Sp Sn 1 02 MeV; • 137 − + . 56 Ba: Sp Sn 4 15 MeV. • 200 − − . 80 Hg: Sp Sn 0 248 MeV. 137 Apart the fourth nucleus, 56 Ba, all nuclei are even A and the values of the separation differences can be interpreted looking at the figure. The nuclei reported in the list increase with A and show a transition from case (a) to case (b). The two assumptions are strictly valid only for the first and last nuclei. The arrow in the figure sketches the transition region.

1.4 Nuclear Decays

Exercise 1.4.1

Denoting by Tα the kinetic energy of the emitted α-particle, we have approximately

A Qα Tα. A − 4

Hence the Qα values are

240 Q × 5.17 5.26 MeV 1 236 240 Q × 5.12 5.21 MeV 2 236

240 236 The γ energy corresponds to the difference between Q1 ( Pu → U) and Q2 (240Pu → 236U∗). Hence we get

Eγ = Q1 − Q2 = 0.05 MeV

Exercise 1.4.2 244 Calling τ1 and N1 the mean lifetime and number of Pu nuclei at time t, τ2 and N2 240 240 the same quantities for U and τ3 and N3 the same quantities for Np, we have τ1 τ2,τ3. Furthermore if t corresponds to the time of the measurement (= 30 d), 2 we have also t τ1. Under these conditions the secular equilibrium equation holds

N1 N2 N3 . τ1 τ2 τ3

2 See e.g. problem 1.4.6. The secular equilibrium is obtained for ω1 ω2,ω3 and ω1t 1. Appendix: Solutions of Exercises and Problems 67

The source has a mass of 1 mol and then N1 = NA, with NA the Avogadro number

( / ) τ T 1 2 14 h × 6.02 · 1023 N(240U) = N = 2 N = 2 N 1.2 · 1013 2 τ 1 (1/2) A 8.1 · 107 × 365 × 24 h 1 T1

( / ) τ T 1 2 N(240 Np) = N = 3 N = 3 N 3 τ 1 (1/2) A 1 T1

67 min × 6.02 · 1023 9.4 · 1011 8.1 · 107 × 365 × 24 × 60 min

The decays involved in the chain are

244 → 240 + α 94 Pu 92 U

240 → 240 + − +¯ν 92 U 93 Np e e

240 → 240 + − +¯ν . 93 Np 94 Pu e e

The activity measured in the α-decay is

23 dN N N ln 2 6.02 · 10 × 0.69 − A = 1 1 = 1 1.6 · 108 s 1. (1/2) 7 dt τ1 8.1 · 10 × 365 × 24 × 3600 s T1

Exercise 1.4.3 The decay constant of 226Ra is

ln 2 0.693 ω = 1.410−11s−1. 3 7 T1/2 1.6 × 10 × 3.15 × 10 s

The activity of a source is given by

dN −ωt −1 A(t) = = ωN(t) = ωN0e [s ] dt

The number of 226Ra nuclei at time 0 is

N 6.02 × 1023 N = A = 2.66 × 1021 0 A 226 Hence we have

−11 21 10 −1 A(0) = ωN0 1.410 × 2.66 10 3.710 s . 68 Appendix: Solutions of Exercises and Problems

Note that this is the deﬁnition of 1 Curie (1 Ci).

Exercise 1.4.4 14 14 → 14 + − +¯ν . The 6 C beta decay is 6 C 7 N e e The specimen activity is given by

14 dN N( C) A = = . dt τ(14C)

14 The number of 6 C nuclei present in the specimen when it was still a living organism is

23 N − 6.02 · 10 N (14C) = f × N (C) = f × m × A 1.3 · 10 12 × 5 × 3.3 · 1011, 0 0 A(C) 12.001

14 where f is the fraction of 6 C nuclei in a living organism, m its mass, NA the ( ) 14 Avogadro number and A C the atomic mass of natural carbon. The 6 C mean 14 14 lifetime is τ( C) = T1/2( C)/ ln 2 8200years. Hence we have for the specimen activity when the organism died

N (14C) 3.3 · 1011 A = 0 1.28 s−1, 0 τ(14C) 8200 × 3.15 · 107 s

The present activity is related to A0 through the equation

14 3600 decays A(t) = A · e−t/τ( C) = 0.5 , 0 2 × 3600 s s hence we get the age of the fossil

A(t) 0.5 T =−τ(14C) × ln −8200 yr × ln 7700 yr. A0 1.28

Exercise 1.4.5 The nucleus 226Ra has a decay constant given by

ln 2 0.693 ω = = 1.410−11 s−1. 3 7 T1/2 1.6 × 10 × 3.15 × 10 s

For 1 g of 226Ra we have then an activity

. × 23 dN NA −11 6 02 10 10 −1 A = = ωN0 = ω × 1 1.410 × 3.710 s . dt A 226

This is the current deﬁnition of 1 Curie (1 Ci). Appendix: Solutions of Exercises and Problems 69

The 60Co source we are considering has an activity of 10 Ci, that is 3.71011 s−1. If m is its mass we get

7 A T / 60 5.26 × 3.15 × 10 m = A 1 2 3.71011 × 8.8mg. 23 NA ln 2 6.02 × 10 0.693

A simpler approach to get the same result is obtained using the following relation, which holds for sources with equal activities

( ) T 1 m1 = A1 × 1/2 . m A (2) 2 2 T1/2

This equation can be used in our case knowing that our source has the same activity of 10 g of 226Ra. Hence we get

(Co) A T / 60 5.26 m = m × Co × 1 2 10 × × 8.7mg. Co Ra A (Cu) 226 1600 Ra T1/2

Exercise 1.4.6 The numbers of nuclei of the three types are ruled by the following nested equations

dN 1 =−ω N dt 1 1 dN 2 = ω N − ω N dt 1 1 2 2 dN 3 = ω N − ω N . dt 2 2 3 3

In our case, the initial conditions are N1(0) = N0, Nk (0) = 0 and dNk /dt(0) = 0 for k = 2, 3. The particular solution for these consitions is

−ω1t N1(t) = N0e ω 1 −ω1t −ω2t N2(t) = N0 (e − e ) ω2 − ω1

−ω −ω −ω e 1t e 2t e 3t N3(t) = N0ω1ω2 + + . (ω2 − ω1)(ω3 − ω1) (ω3 − ω2)(ω1 − ω2) (ω1 − ω3)(ω2 − ω3) uhe nucleus 3 is stable and then ω3 = 0 and N3(t) can be written as

−ω −ω e 1t e 2t N3(t) = N0 1 + + . ω1/ω2 − 1 ω2/ω1 − 1 70 Appendix: Solutions of Exercises and Problems

N1 N3 ω -1 0.8 1 = 10 s ω -1 2 = 50 s ω = 0 0.6 3

0.4

0.2 N2

0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 time [s]

Fig. 1.3 Relative abundances for a decay chain with three nuclei having the decay constants given in the ﬁgure

In Fig. 1.3 the three nuclear populations are shown as a function of time for the decay = / constants given in the text. For t 1 4 s we obtain −ω −ω + e 1t + e 2t 1 ω /ω − ω /ω − N3 = 1 2 1 2 1 1 . . −ω t 10 9 N1 e 1

Exercise 1.4.7 The fraction of 238U isotopes decayed in 2.5 109 years is t t ln 2 f = 1 − exp − = 1 − exp − τ T / 1 2 2.5109 × 0.693 1 − exp − 32% 4.5109

The speciﬁc activity is the activity per unit mass. Hence we have A N ln 2 a = = A M A T1/2

6.02 1023 0.693 μCi 1.23 104 s−1 · g−1 0.33 238 4.5109 × 3.15 107 g

Exercise 1.4.8 44 → 40 α. 1. 22Ti 20Ca + Appendix: Solutions of Exercises and Problems 71

This decay is not allowed. Only nuclei having A ≥ 200 can fulﬁll the kinematical conditions for the α-decay. 241 → 237 α. 2. 95 Am 93 Np + This decay is allowed if Qα > 0. Using the SEMF we obtain

Qα = M(241, 95) − M(237, 93) − Mα = B(237, 93) − B(241, 95) + Bα

= 1798 − 1820 + 28.3 5.91 MeV.

Hence the decay is allowed. 141 → 141 + + ν . 3. 55 Cs 56 Ba + e e This decay is forbidden because charge is not conserved. 69 → 69 − +¯ν . 4. 28Ni 29Cu + e e This decay is allowed, provided that we have Qβ− > 0.

Qβ− = M(69, 28) − M(69, 29) − me

= 28Mp + 41Mn − B(69, 28) − 29Mp − 40Mn + B(69, 29) − me

= Mn − Mp − me + B(69, 29) − B(69, 28) 0.782 + 600.0 − 593.5 7.3MeV

Hence the decay is allowed. Exercise 1.4.9 As in problem 1.4.6, the time evolution of the nuclei involved in the chain is

−ω1t N1(t) = N0e ω 1 −ω1t −ω2t N2(t) = N0 (e − e ) ω2 − ω1

−ω −ω −ω e 1t e 2t e 3t N3(t) = N0ω1ω2 + + . (ω2 − ω1)(ω3 − ω1) (ω3 − ω2)(ω1 − ω2) (ω1 − ω3)(ω2 − ω3)

−1 In our case the third equation is not used. We have ω1 = ln 2/2.25 = 0.31 min , −1 ω2 = ln 2/22.9 = 0.03 min . The maximum N2 is found solving the equation dN 2 = 0 dt

−ω1t −ω2t −ω1 e + ω2 e = 0 whose solution is ln(ω /ω ) t = 1 2 8.5min ω1 − ω2

The time dependence of the nuclear fractions is shown below. 72 Appendix: Solutions of Exercises and Problems

0.9 79Kr 0.8 79Rb 0.7

0.6

0.5

0.4 79Sr 0.3

0.2

0.1

0 0 102030405060708090100 time [min]

Exercise 1.4.10 238 222 Denoting by NU and NRn the numbers of U and Rn nuclei, ωU and ωRn their decay constants, the condition of secular equilibrium can be written as

NU ωU = NRn ωRn.

ωRn is related to the speciﬁc activity measurement as

A N ω N ω a = = Rn Rn = U U V V V

3 238 where V is the volume of the basement (60 m ). NU can be expressed as the U concentration ρU times the volume from which the Radon gas diffuses

ρ Sdω a = U U , V where S is the surface of the walls (94 m2) and d is the thickness (0.02 m) of the layer from which the gas diffuses. Hence we have

/ aV aVT1 2 100 × 60 × 4.5109 × 3.15 107 ρ = = U 6.5 × 1020 m−3 U 3 SdωU Sd ln 2 94 × 0.02 × 0.693 m

Exercise 1.4.11

The Qα-value of the decay can be obtained from the alpha decay energy

A 239 Qα = Tα = 5.144 5.232 MeV. A − 4 235 Appendix: Solutions of Exercises and Problems 73

The measured power is equal to the intensity of the alpha decays multiplied by the released energy (Qα). Hence we get

W 0.231 Joule/s 0.231 Joule/s = . 11 −1. I − 2 76 10 s Qα 5.232 106 eV 5.232 106 1.610 19Joule

The half-life is then

239 239 N( Pu) ln 2 NA m( Pu) ln 2 T / = = 1 2 I AI

6.02 1023 × 120 × 0.693 7.57 1011 s−1 24000 yr. 239 × 2.76 1011

1.5 Nuclear Models

Exercise 1.5.1 The Saxon-Woods potential has the following expression

−V V (r) = 0 , + r−R 1 exp d where −V0, R and d are the three potential parameters, representing respectively the minimum depth, the nuclear radius and the thickness of region where nuclear matter vanishes. Taken any spherical potential well, a larger radius generates eigenfunctions which are contained in larger volumes. As a consequence the energy levels (i.e. the eigen- values) decrease. Hence for a nucleus with larger radius we expect lower energy levels. A more quantitative result cannot be obtained, as the radial Schödinger equation for a Saxon-Woods potential is not analytically integrable. However using the Fermi gas model we can estimate the relative effect. The Fermi energy is:

/ 1 2 9π 2 3 EF = 33MeV, (1.14) 2m r0 8 where m is the nucleon mass and r0 1.2 fm is the coefﬁcient of the nuclear radius 1/3 A-dependence (R r0 A ). This energy represents the maximum kinetic energy of nucleons in the nucleus. The energy of the ground state is obtained as EGS = −V0 + EF , with V0 ≈ 41 MeV to agree with a binding energy per nucleon of about 8MeV.From(1.14) increasing by 50% the nuclear radius one gets

EGS =−V0 + EF (1.5 r0) −41 + 15 −26 MeV, 74 Appendix: Solutions of Exercises and Problems which has to be compared with -8 MeV for the standard radius. The binding energy is related to the ground state energy (about equal to its absolute value), hence it increases with the nuclear radius.

Exercise 1.5.2 The carbon isotopes have 6 protons. These are all contained in fully closed shells 2 4 according to the conﬁguration (1s1/2) (1p3/2) . Hence they do not contribute to the spin-parities of the nuclei. These are instead determined by the last neutron shells. The conﬁgurations of the carbon isotopes are 11 2 3 C:(1s1/2) (1p3/2) 12 2 4 C:(1s1/2) (1p3/2) 13 2 4 1 C:(1s1/2) (1p3/2) (1p1/2) 14 2 4 2 C:(1s1/2) (1p3/2) (1p1/2) In all cases the last shell has l = 1. We then have for odd-N isotopes 11C:J = 3/2, P = (−1)1 =− ⇒J P = 3/2− 13C:J = 1/2, P = (−1)1 =− ⇒J P = 1/2−. Instead for even-N isotopes, all neutrons are paired and then 12C, 14C:J = 0, P =+ ⇒J P = 0+

Exercise 1.5.3 The ﬁrst two nuclei are odd-A. Hence their spin and parity is determined by the last unpaired nucleon. The shell conﬁgurations are 33 : ( )2( )4( )2( )6( )2( )1 16S n 1s1/2 1p3/2 1p1/2 1d5/2 2s1/2 1d3/2 39 : ( )2( )4( )2( )6( )2( )3 19K p 1s1/2 1p3/2 1p1/2 1d5/2 2s1/2 1d3/2 Both nucleons have l = 2. Their spin and parity are 33 : = = (− )2 = ⇒ P = / + 16S n J 3/2, P 1 + J 3 2 39 : = = (− )2 =+ ⇒ P = / + 19K p J 3/2, P 1 J 3 2 64 P = + 28Ni is an even-even nucleus, all nucleons are paired and then J 0 . Exercise 1.5.4 Using the Fermi gas distribution we have p 2 p F p d3 p 4π F p4dp 3 p2 E = 0 2M = 0 = F k pF 3 pF π 2 0 d p 2M 0 4 p dp 5 2M

1/3 where pF = (9π) is the Fermi momentum (r0 = 1.2 fm) and M is the nucleon 2r0 mass (can be assumed equal for the purpose). Multiplying and dividing by c2,we obtain 3(c)2(9π)2/3 3 × 1972 × 28.272/3 E = 20 MeV. k 2 2 × . 2 × 40 r0 Mc 40 1 2 940

This expression does not depend on the content of protons (Z) and neutrons (N). Therefore the mean kinetic energy is the same for all nuclei. Appendix: Solutions of Exercises and Problems 75

Exercise 1.5.5 The nuclear shells involved and the spin-parity of the ground states are − 15 2 4 1 P 1 : ( / ) ( / ) ( / ) ⇒ = – 7 N odd-A nucleus, p 1s1 2 1p3 2 1p1 2 J 2 + 27 2 4 2 6 1 P 1 : ( / ) ( / ) ( / ) ( / ) ( / ) ⇒ = – 12Mg odd-A nucleus, n 1s1 2 1p3 2 1p1 2 1d5 2 2s1 2 J 2 60 ⇒ P = + – 28Ni even-even nucleus J 0 87 : ( )2( )4( )2( )6( )2( )4( )8 – 38Sr odd-A nucleus, n 1s1/2 1p3/2 1p1/2 1d5/2 2s1/2 1d3/2 1 f7/2 + 4 6 2 9 P 9 ( / ) ( / ) ( / ) ( / ) ⇒ = 2p3 2 1 f5 2 2p1 2 1g9 2 J 2 Exercise 1.5.6 All these isotopes can be unstable because of beta decay. Gamma decay is not possible because they are in the ground states. Alpha decay is kinematically forbidden for A 200. To establish if they are stable it is then necessary to evaluate their Qβ values. 15 β− β+ 8 O nuclide. We have to calculate Q−, Q+ and Q EC respectively for , decays and electron capture (EC). We have

Q− =−B− + 0.782 MeV

Q+ =−B+ − 1.804 MeV

Q EC = Q+ + 1.022 MeV where B∓ = B(A, Z) − B(A, Z ± 1), the difference between the binding energies of the parent and daughter nuclei, can be derived from the SEMF. < = = 15 We obtain Q− 0, instead Q+ 2.44 MeV and Q EC 3.46 MeV. 8 O is then unstable and can decay by both β+-decay and EC. This isotope is odd-A, so the spin and parity are determined by the unpaired neutron of the last shell. The neutron shell conﬁguration and spin-parity are 2 4 1 P − n : (1s1/2) (1p3/2) (1p1/2) ⇒ J = 1/2 . In the shell model the magnetic moment is μ = gJ J. We have an unpaired neutron (gl = 0, gs =−3.83 n.m.) and gJ is given by

j( j + 1) − l(l + 1) + s(s + 1) g = g (1.15) J s 2 j( j + 1)

15 = which for 8 O turns out to be gJ 1.28 corresponding to a magnetic moment μ 0.64 n.m. 16 8 O nuclide. This nucleus is even-even. Furthermore it is the well-known stable and most abundant oxygen isotope. Hence we have J P = 0+ and μ = 0. 17 8 O nuclide. Evaluating Q−, Q+ e Q EC as for the ﬁrst nuclide we obtain negative 17 values for all of them. Hence 8 O is stable. The same nuclide is odd-A. Again the spin and parity are determined by the unpaired neutron of the last shell. The shell conﬁguration and spin-parity are 2 4 2 1 P + n : (1s1/2) (1p3/2) (1p1/2) (1d5/2) ⇒ J = 5/2 . 76 Appendix: Solutions of Exercises and Problems

Using Eq. (1.15) we now get gJ =−0.77 corresponding to a magnetic moment μ −1.92 n.m.

Exercise 1.5.7 We have odd-A nuclei and so the spin and parity of the ground states are that of the unpaired nucleon. The maximum occupation is 15, corresponding to the neutron 29 number for 14Si. The shell sequence up to 20 in the standard shell model, that is with inverse spin-orbit coupling, is

2 4 2 6 2 4 (a)(1s1/2) (1p3/2) (1p1/2) (1d5/2) (2s1/2) (1d3/2) .

If instead the spin-orbit coupling were direct (case b) the shell sequence proceeds with increasing J values as

2 2 4 4 2 6 (b)(1s1/2) (1p1/2) (1p3/2) (1d3/2) (2s1/2) (1d5/2) .

7 For 3Li, whose unpaired nucleon is a proton, we have

2 1 P − (a) p : (1s1/2) (1p3/2) J = 3/2

2 1 P − (b) p : (1s1/2) (1p1/2) J = 1/2 .

29 For 14Si a neutron is unpaired and we have

2 4 2 6 1 P + (a) n : (1s1/2) (1p3/2) (1p1/2) (1d5/2) (2s1/2) J = 1/2

2 2 4 4 2 1 P + (b) n : (1s1/2) (1p1/2) (1p3/2) (1d3/2) (2s1/2) (1d5/2) J = 5/2

Exercise 1.5.8 52Cr is even-even and then spin-parity is J P = 0+. The other Cr isotopes are odd-A and J P is that of the unpaired nucleon. The proton number is even, hence only the neutron shell conﬁguration is relevant. We have 51 2 4 2 6 2 4 7 Cr 27 n : (1s1/2) (1p3/2) (1p1/2) (1d5/2) (2s1/2) (1d3/2) (1 f7/2) 55 2 4 2 6 2 4 8 3 Cr 31 n : (1s1/2) (1p3/2) (1p1/2) (1d5/2) (2s1/2) (1d3/2) (1 f7/2) (2p3/2) . Therefore spin and parity of the ground states are 51Cr J = 7/2, P = (−1)3 ⇒ J P = 7/2− 55Cr J = 3/2, P = (−1)1 ⇒ J P = 3/2− These two isotopes ate unstable because of β decay. To ﬁnd the possible decay − + modes we calculate Q−, Q+ e Q EC respectively for β , β and electronic capture (EC). These are Q− =−B− + 0.782 MeV

Q+ =−B+ − 1.804 MeV Appendix: Solutions of Exercises and Problems 77

Q EC = Q+ + 1.022 MeV, where B∓ = B(A, Z) − B(A, Z ± 1) is the binding energy difference corresponding to each decay. To get B∓ we use the SEMF. 51 . − For the 24Cr isotope, Q+ and Q− are both negative, yet we ﬁnd Q EC 1 52 . + . . 51 51 1 804 1 022 0 74 MeV. This means that 24Cr transmutes to 23V by electron capture. 55 . + . . For 24Cr, Q+ and Q EC are both negative, but Q− 1 19 0 782 1 97 MeV. 55 β− Hence this nucleus decays to 25Mn by -decay. Exercise 1.5.9 (a) 57Cu e 57Ni are mirror nuclei with a single nucleon (valence nucleon) out of complete shells. The valence nucleon is a proton for 57Cu and a neutron for 57Ni. The shell sequence is

1s1/2 1p3/2 1p1/2 1d5/2 2s1/2 1d3/2 1 f7/2 2p3/2 1 f5/2 ...

The shell conﬁguration up to the valence nucleon (occupancy no. 29) is

2 4 2 6 2 4 8 1 (1s1/2) (1p3/2) (1p1/2) (1d5/2) (2s1/2) (1d3/2) (1 f7/2) (2p3/2) and the ﬁrst excited level corresponds to the following shell, 1 f5/2. Hence we have for spin and parity

3 3 − 5 5 − GS : l = 1, j = ⇒ J P = ; 1st Exc : l = 3, j = ⇒ J P = . 2 2 2 2

(b) The magnetic moment is μ = g j j, where

j( j + 1) + l(l + 1) − s(s + 1) j( j + 1) − l(l + 1) + s(s + 1) g = g + g j l 2 j( j + 1) s 2 j( j + 1)

For j = l + 1/2, which holds for both nuclei since the valence nucleus is in p3/2, the previous equation simpliﬁes to

jgj = gll + gs/2

57 For Cu, substituting the orbital and spin g-factors for a proton, gl = 1, gs =+5.6 n.m., we obtain

57 μ( Cu) = jgj = 1 × 1 + 5.6/2 = 3.8n.m.

57 For Ni, having a valence neutron, the g-factors are gl = 0, gs =−3.8 n.m. and we have 57 μ( Ni) = jgj = 0 × 1 − 3.8/2 =−1.9n.m. 78 Appendix: Solutions of Exercises and Problems

2 (c) Qβ+ is given by [M(A, Z) − M(A, Z − 1) − 2m]c , where M denotes the atomic mass and m the electron mass. Since the parent (57Cu) and daughter (57Ni) nuclei are mirror nuclei, the binding energy difference is only due to the difference in the Coulomb energies. To write the atomic mass difference we have only to subtract3 the mass difference because a proton is exchanged into a neutron after the decay. Hence we can write

2 2 Mc ≈ Ec(Z) − Ec(Z − 1) + (Mp − Mn)c

3 e2 3 αc 2Z + 1 [Z2 − (Z − 1)2]+(M − M )c2 + (M − M )c2 p n 1/3 p n 5 4π0 R 5 r0 A

3 197 59 + 938.27 − 939.57 9.72 MeV 5 137 × 1.2 571/3

The maximum positron energy is equal to Qβ+ and then we have

2 Tmax = Qβ+ = (M − 2m)c ≈ 9.72 − 2 × 0.511 8.7MeV.

It is worth to notice that using the SEMF the result is 8.5 MeV.

Exercise 1.5.10

The shell sequence up to 14 is 1s1/2 1p3/2 1p1/2 1d5/2. 17 The spin and parity of 8 O is that of the uncomplete neutron shell: + 17 1 5 P 5 − : ( / ) = , = ⇒ = 8 O, n 1d5 2 l 2 j 2 J 2 . 18 In the case of 9 F, there are valence nucleons in both proton and neutron shells. Hence the shell model prediction is not unique. The valence shell is the same 1d5/2.The 5 ⊕ 5 resulting spin comes from the angular momentum composition 2 2 , whereas the parity is the product (−1)2 × (−1)2 =+1. So we have − 18 : ( )1 ( )1 ⇒ P = +, + +, +, +, + 9 F, p 1d5/2 ,n: 1d5/2 J 0 1 2 3 4 5 . (From measurements we have J P = 1+). For the last nucleus we need to extend the shell sequence. The two last shells up to 207 an occupation 82 are 2d3/2 3s1/2, instead up to 126 are 1i13/2 3p1/2. 82 Pb125 has a valence neutron in the 3p1/2 shell. Hence we ﬁnd − 207 1 1 P 1 − : ( / ) = , = ⇒ = 82 Pb, n 3p1 2 l 1 j 2 J 2 .

3In the SEMF the energy term and the mass term have opposite signs. Appendix: Solutions of Exercises and Problems 79

A.2 Solutions of Particle Physics (Chapter 2)

2.1 Fundamental Interactions

Exercise 2.1.1

Exercise 2.1.2 Let us ﬁrst convert the cross section from√ natural to CGS units. In natural units = . × −5 −2 2 = . × G F 1 2 10 GeV ; then expressing s in GeV, we obtain G F s 1 44 √ 2 −10 s −2 10 1GeV GeV . To perform the conversion we use the relationship c 197 MeV·fm 1.97 × 10−14 GeV·cm, which allows to get 1/GeV = 1.97 × − 10 14 cm. Hence √ s 2 G2 s = 5.6 × 10−38 cm2. F 1GeV

2 The CMS square total energy of the ν-nucleon is given by the invariant (pν + pp) , where pν and pp are the 4-momenta of the neutrino and proton respectively. There- = 2 + 2 + = 2 = . fore s Mp Mν 2Mp Eν : substituting Mν 0 and neglecting Mp (Mp 0 94 GeV/c2) in the high energy limit, we get Eν 2 s 2M Eν 1.88 GeV p 1GeV and for the cross section −38 2 × 5.610 × 1.88 Eν Eν σ 7.4 × 10−39 cm2. 28.27 1GeV 1GeV

The number of scatterers (nucleons) per unit volume is n = ρ/Mp = NAρ 1.3 × 1024 cm−3 and hence the interaction length is 80 Appendix: Solutions of Exercises and Problems −1 1 Eν λ = 1.0 × 1014 cm. σ n 1GeV

An estimate of the ν energy above which the Earth becomes opaque is obtained equating such length to the Earth diameter D = 1.2 × 109 cm. This energy turns out 4 to be Eν > 8.3 × 10 GeV.

Exercise 2.1.3 • e+ + e− → μ+ + μ− : γ + Z 0 − • n → p + e +¯νe : W − − • μ → e +¯νe + νμ : W − − 0 • νe + e → νe + e : W + Z − − 0 • νμ + e → νμ + e : Z

Exercise 2.1.4 All the processes are allowed, except p + p → K + + p, which is forbidden because of baryon conservation (Bini = 2 = Bfin = 1) and strangeness conservation (Sini = 0 = Sfin =+1). The first two processes

γ + γ → γ + γ, e+ + e− → 4γ are due to electromagnetic interaction, the third and the ﬁfth ones

− − − p +¯p → W + X,νμ + e → νμ + e to weak interaction. The Feynman diagrams of the allowed reactions are shown below. Appendix: Solutions of Exercises and Problems 81

Exercise 2.1.5

• e+ + e− → γ + γ : allowed—e.m. interaction • π − + n → K − + : forbidden—strangeness not conserved (K − = su¯,= uds ⇒ Sini = 0 = Sfin =−2). + + + • → n + e + νe: forbidden—weak interaction but two flavors changed ( = uus, n = udd). + + + • → + e + νe: allowed—weak interaction(u → d + W ). 0 + − • ρ → K + K : forbidden—kinematics (mfin > mini). − − − 0 •¯νe + e →¯νe + e : allowed—weak interaction (W + Z ) − − − 0 • νe + e → νe + e : allowed—weak interaction (W + Z ). The Feynman diagrams for the allowed processes are shown below.

Exercise 2.1.6 a. π − + p → 0 + K 0: it is a strong interaction process. X must have Q = 0, B = 0 and strangeness S =+1(s¯) (because 0 = uds and then S =−1). K 0 = ds¯ possesses all these features. + b. e + n → p +¯νe: it is a weak process. X must have Q = 0 and electron lepton number Le =−1. Hence it is an ν¯e. The same result can be obtained using the neutron − beta decay, n → p + e +¯νe, and moving the electron to the initial state. 82 Appendix: Solutions of Exercises and Problems c. 0 → + X. The missing particle must be a meson. Looking at the particles involved (all hadrons) the decay can be either strong or weak.4 If the decay were strong, the meson should have Q = 0 and strangeness S =−1. It might be a K¯ 0, but the system + K¯ 0 is too heavy for the 0 decay. So a strong decay is excluded. The decay is weak and the strangeness conservation is not holding any more, being replaced by S =±1: a neutral pion is the right answer. Hence it is 0 → + π 0. Exercise 2.1.7 The ﬁgure below shows the Feynman diagrams for all the processes. It has to be noticed that (a) and (b) can also occur as neutral current processes (Z 0 instead of γ ). This concurrence is more and more important as energy increases.

Exercise 2.1.8 Hereafter is the list of reactions (A for allowed, F for forbidden, “conservation” is implicit): + + 1. μ → e + γ : F, violates Le and Lμ. − 2. e → νe + γ : F, violates charge. 3. p + p → + + K +: F, violates B. 4. e+ + e− → γ : F, violates energy. + 5. νμ + p → μ + n: F, violates Lμ. − 6. νμ + n → μ + p: A, see ﬁgure.

4For the moment we disregard the knowledge that all the J P = (1/2)+ baryons, except p and 0, decay weakly. Appendix: Solutions of Exercises and Problems 83

+ 7. e + n → p + νe: F, violates Le − 8. e + p → n + νe: A, see ﬁgure. + 0 + 9. π → π + e + νe: A, see ﬁgure for one of the possible graphs 10. p +¯p → Z 0 + X: A, a possible case is shown, with qq¯ fragmentations omitted.

2.2 Hadrons

Exercise 2.2.1 (1) K 0-mesons as in (2.1) are produced in a strong interaction process. The following quantities are then conserved: electric charge Q, baryon number B, lepton number L, strangeness S. Considering the initial state and the K 0 in the final state, the requirements for X are Q =+2, B =+2, L = 0, S =−1. No known particle exists with such numbers. The minimum number of particles composing X is two because two baryons can realize a system with B =+2. Hereafter a few processes fulfilling these requirements are listed: • p + p → K 0 + p + + • p + p → K 0 + + + + • p + p → K 0 + 0 + ++ • p + p → K 0 + K¯0 + p + p • p + p → K 0 + K¯0 + p + n + π + • p + p → K 0 + K 0 + + + + • p + p → K 0 + 0 + + + π + • p + p → K 0 + K 0 + 0 + 0 + π + + π + We further notice that the minimum energy (threshold energy) is different for each of the listed processes. (2) Several experimental set-ups can be used to study reaction (2.1), depending on the quantities to be measured and the particle identification required. 84 Appendix: Solutions of Exercises and Problems

Let us assume that the experimental configuration consists in a beam of protons hitting a fixed target. Since we want to select events including K 0-mesons, the observation of their decays is mandatory. As K 0 is neutral, its decay is detected through the observation of the decay particles. One can use various types of detectors posi- tioned downstream of the target (e.g., wire or drift chambers) or imaging detectors acting as target as well (e.g., bubble chamber). With such detectors it is possible to observe the decay into π +π − from their tracks. To measure charge and momentum of the pions a suitable magnetic field is the best solution. The detection of the decay into neutral pions is much more challenging, because it requires the observation of the two photons emerging from the quasi immediate π 0 decay. This can be achieved with a downstream electromagnetic calorimeter or, in the case of a bubble chamber, using a heavy liquid filling (e.g. freon). 0 −t/τ The K decay follows the exponential decay law N(t) = N0e , where N(t) (N0) is the number of particles at time t (time 0) and τ is the mean lifetime. It can be reasonable to require that 99% of the neutral kaons decay in the detector. This requirement determines the size of the experimental set-up. We have T dt N(t) 0.99 = = 1 − eT/τ 0 τ N0 hence T 4.6 τ. Therefore the minimum length of the experimental set-up is p L 4.6 βγcτ = 4.6 cτ 0.74 m. m Exercise 2.2.2 All the reactions are strong interaction processes. Considering the particles involved we need to check the conservation of the following quantities: electric charge Q, baryon number B and strangeness S.

• K − + p → − + K + + K 0: allowed; • ψ → π + + π 0 + π −: allowed; • π − + p → + + K −: forbidden for S non conservation; • π − + p → π 0 + π 0: forbidden for B non conservation; • p + p → n + ++ + p +¯p: allowed.

Exercise 2.2.3 The answers about the decays and the interaction type are • φ → ρ0 + π 0: allowed, strong interaction; • π 0 → e+ + e− + γ : allowed, e.m. interaction; − 0 − • → + μ +¯νe: forbidden, violates the electron and muon numbers con- servations; • − → n + π −: allowed, weak interaction; • − → π 0 + π −: forbidden, violates the baryon number conservation. Appendix: Solutions of Exercises and Problems 85

Exercise 2.2.4 To get the e+ + e− → μ+ + μ− cross section in cm2 we simply multiply it by (c)2: 4πα2 43.14 0.197 GeV · fm 2 GeV2 σ(μ+μ−) = (c)2 × 86.6nb 3s 3s 137 s

Neglecting strong interaction effects, the cross section into hadrons can be estimated from the ratio R σ(hadrons) R = = C Q2 σ(μ+μ−) q q where C is the number of quark colors (3), Qq is the charge√ of the quark q (in e units) and the sum includes those quarks for which m(qq¯)< s.At2GeVu, d e s fulﬁll such condition and then 1 4 1 GeV2 σ(hadrons) = 3 × + + × 86.6nb 43.3nb 9 9 9 4GeV2

Exercise 2.2.5 We have c 197 MeV fm −21 τ /ψ = = 7.2 × 10 s J (J/ψ) (J/ψ)c 0.091 MeV 3 1023 fm/s

The decay time corresponds to a strong interaction decay.

Exercise 2.2.6 The beam energy is above the energy threshold for the production of strange particles, but below that for producing particles with heavier quarks. Therefore the simplest hypothesis for the event is the associated production of and K 0 observed through their respective decays into p + π − and π + + π −. Having in mind also the two charged tracks, the simplest interpretation for the event is

π + + p → π + + π + + + K 0

To verify the correctness of the interpretation and to assign a speciﬁc particle to each V0, we assume that the negative track is a π −, whereas the positive one can be either p (-hypothesis) or π + (K 0-hypothesis). 0 LetuscallV1 the ﬁrst vertex. If it is a decay, we have 2 = 2 + 2 + 2 + 2 2 + 2 − θ = M m p mπ 2 p1+ m p p1− mπ 2p1+ p1− cos 1

= 0.9382 + 0.1392 + 2 × 1.02 × 1.905 − 2 × 0.4 × 1.9 × cos 24.5◦ 3.40 GeV2 86 Appendix: Solutions of Exercises and Problems hence M 1.84 GeV, which is inconsistent with the hypothesis, since it differs by more than 5% from the mass (1.116 GeV/c2). 0 0 If V1 is a K decay, we have 2 = 2 + 2 + 2 + 2 2 + 2 − θ = M mπ mπ 2 p1+ mπ p1− mπ 2p1+ p1− cos 1

◦ = 0.1392 + 0.1392 + 2 × 0.423 × 1.905 − 2 × 0.4 × 1.9 × cos 24.5 0.267 GeV2 hence M 0.517 GeV, which is consistent with the hypothesis, being within 5% from the K 0 mass (0.498 GeV/c2). 0 V2 is the second vertex. If it is a decay, we have 2 = 2 + 2 + 2 + 2 2 + 2 − θ = M m p mπ 2 p2+ m p p2− mπ 2p2+ p2− cos 2

= 0.9382 + 0.1392 + 2 × 1.20 × 0.286 − 2 × 0.75 × 0.25 × cos 22◦ 1.24 GeV2 hence M 1.11 GeV, which differs from the mass by less than 5%. 0 To further conﬁrm the -hypothesis for V2, we calculate the invariant mass for a K 0 as 2 = 2 + 2 + 2 + 2 2 + 2 − θ = M mπ mπ 2 p2+ mπ p2− mπ 2p2+ p2− cos 2

= 0.1392 + 0.1392 + 2×0.76 × 0.286 − 2 × 0.75 × 0.25 × cos 22◦ 0.126 GeV2

M 0.354 GeV is inconsistent with the K 0 mass. 0 0 0 As a conclusion V1 is a K ,V2 is a . The lifetime of each particle is l l m t = = × βγc c p where m and p are mass and momentum of the decaying particle. We have = 2 + 2 + θ pK 0 p1+ p1− 2p1+ p1− cos 1

0.42 + 1.92 + 2 × 0.4 × 1.9 × cos 24.5◦ 2.27 GeV/c = 2 + 2 + θ p p2+ p2− 2p2+ p2− cos 2

0.752 + 0.252 + 2 × 0.75 × 0.25 × cos 22◦ 0.99 GeV/c and the lifetimes are Appendix: Solutions of Exercises and Problems 87 . 37 cm 0 498 −10 t 0 × 2.710 s K 3 × 1010 cm/s 2.27 . 11 cm 1 116 −10 t × 4.110 s 3 × 1010 cm/s 0.99

Exercise 2.2.7 a. Forbidden: strangeness is not conserved. b. Forbidden: electric charge is not conserved. c. Allowed. d. Forbidden: energy is not conserved. e. Forbidden: strangeness is not conserved. f. Allowed.

Exercise 2.2.8 a. 0 decays by electromagnetic interaction. For this interaction the quark ﬂavor is conserved as for the strong interaction. The strangeness conserving decay is possible because a lighter baryon with the same strangeness does exist. The 0 → decay would be also possible by strong interaction if accompanied by π 0, but there is not enough energy [M(0)

Exercise 2.2.9 0 (a) Denoting by σ and p the spin and momentum, and by pK the K momentum, the vector product t = σ × ( p × pK ) is parallel to the scattering plane and proportional to the spin value. Hence it is proportional to the component of the spin in this plane. t is an axial vector and then must be zero if parity is conserved (t →−t under parity transformation). This is the case for the strong reaction π − + p → + K 0. Being null the spin component in the scattering plane, the spin can only be normal to this plane. (b) Using the star superscript for center-of-momentum system (CMS) kinematic variables, we have the following relations 2 2 2 2 π = pπ + mπ = pπ + mπ 1.01 GeV/c ∗ = ∗ + ∗ = 2 + 2 + . E π p mπ m p 2m p π 1 67 GeV 88 Appendix: Solutions of Exercises and Problems

βCM =|pπ |/(π + p) = pπ /(π + m p) 0.513

∗ γCM = (π + p)/E 1.16

∗2 ∗2 [E − (m + m )2][E − (m − m )2] p∗ =|p∗|= K K 0.203 GeV/c 2E∗ ∗ ∗2 2 = p + m 1.13 GeV.

θ = 0 in the Laboratory system (LS) corresponds to θ ∗ = 0 in the CMS. Hence the momentum in the LS is

∗ ∗ p = γCM (p + βCM · ) 0.915 GeV/c

The mean decay path of the -particle is

p λ = cτ · βγ = cτ · 6.47 cm m and the probability that it decays before reaching the detector is 1 L P(< L) = exp(−l/λ) dl 1 − exp(−10/6.47) 79% λ 0

L L φ = ωt = ω = ω , v βc where ω is the Larmor angular frequency μ ω = B .

Hence we have

μ . × . −14 / × × . 2 + . 2 φ = BL E 0 61 3 15 10 MeV T 20 T 10 cm × 0 915 1 116 − c p 197 10 13 MeV cm 0.915

0.308 rad 17.6◦

(d) The decay asymmetry is deﬁned as 1 ( θ ∗) θ ∗ ( − α 2/ )|1 α 0 N cos d cos x x 2 0 1 f+ = = = 1 − . 1 ∗ ∗ 2 1 ( θ ) θ (x − αx /2)|− 2 2 −1 N cos d cos 1 Appendix: Solutions of Exercises and Problems 89

Thus we have for the asymmetry parameter α:

α = 2(1 − 2 f+) 0.72

(e) The decay asymmetry is a consequence of the parity non conservation in weak interactions, as for the decay → π − + p. In fact we have N(θ ∗) = N(π − θ ∗).

Exercise 2.2.10 The event in the text is interpreted as π − + p → + K 0. From momentum conser- = + vation, pπ p pK , we derive the momentum = − p pπ pK whose absolute value is = 2 + 2 − | || | θ p pπ pK 2 pπ pK cos K

1.52 + 0.522 − 2 × 1.5 × 0.52 × cos 58◦ 1.3GeV/c a. Assuming that the particles decayed from the second V0 are a proton (with momentum p+) and a negative pion (with momentum p−), the square of the invariant mass is 2 = 2 + 2 + 2 + 2 2 + 2 − (θ + θ ) = M m p mπ 2 p+ m p p− mπ 2p+ p− cos + −

= 0.9382 + 0.1402 + 2 × 1.31 × 0.25 − 2 × 0.92 × 0.21 × cos 18◦ 1.195 GeV2 √ Hence the invariant mass is 1.195 1.09 GeV which does not correspond to a -particle. The invariant mass is smaller: this implies that (at least) a neutral particle is not observed in the decay (as hypothesized in b.). This fact can be put in evidence using the momentum conservation in the longitudinal direction (i.e. along the momentum). The total longitudinal momentum of the decay products is

◦ ◦ p+ cos θ+ + p− cos θ− = 0.92 cos 4 + 0.21 cos 14 1.12 GeV/c √ which is smaller than the momentum (1.3 GeV/c) by more than 5% 2. Finally we notice that the text did not provide the azimuthal angles of the decay products. Having these angles it would have been possible to evaluate the momentum vectors of these particles. The best way to verify the presence of unobserved neutrals is showing that p, p+ and p− are not lying in the same plane or equivalently the sum of the proton and pion transverse momenta is not zero. b. If a neutrino is the missing neutral particle, its longitudinal momentum is

(pν )L = p − (p+ cos θ+ + p− cos θ−) 1.3 − 1.12 = 0.18 GeV/c c. The lifetime is 90 Appendix: Solutions of Exercises and Problems

l l m 10 cm 1.116 t = = 2.04 10−10 s. βγc c p 31010 cm/s 1.3

Exercise 2.2.11 (a) In the quark model baryons are 3-quark systems. Since quarks are fermions with spin 1/2, baryons must have a half-integer spin. (b) An antibaryon is constituted of 3 antiquarks whose charges are either −2/3or +1/3. Hence the maximum charge is +1[=3×(+1/3)]. (c) A meson is a quark-antiquark system. To get S =−1, the quark must be s whose charge is Qq =−1/3. It follows that the charge of the meson can be either −1 (Qq¯ =−2/3) or 0 (Qq¯ =+1/3). Exercise 2.2.12 (a) Mesons are qq¯, the charges are +2/3 and −1/3forq and −2/3 and +1/3forq¯. Combining the four possible cases, one ﬁnds that the charges for mesons are −1, 0 and +1. (b) Antibaryons are q¯q¯q¯. Again there are four possible cases which are −2, −1, 0, +1.

2.3 Weak and Electro-Weak Interactions

Exercise 2.3.1 The neutrino mean free path in Iron is

λ = 1 n pσν where n p = ρFe/m p is the number of nucleons per unit volume. We have

7.9g/cm3 n 4.7 × 1024 cm−3 λ = 7.1 × 1010 cm p 1.67 10−24 g

Then, if f = 1/109 is the fraction of interacting neutrinos, the corresponding thickness is L = f λ = 71 cm

Exercise 2.3.2 For an estimate of the branching ratios we assume that they are simply proportional to the transition rates as given by the Fermi golden rule. Hence we have

( 0 → − +ν ) |M( 0 → − +ν )|2 ρ( 0 → − +ν ) BR D K e e D K e e × D K e e , 0 − + 0 − + 2 0 − + BR(D → π e νe) |M(D → π e νe)| ρ(D → π e νe) Appendix: Solutions of Exercises and Problems 91 where M denotes the transition amplitude and ρ the phase space factor. In the ﬁrst ratio all the terms cancel but the effective coupling constants. These are gw cos θC 0 − + 0 − + for D → K (c → s + W ) and gw sin θC for D → π ,(c → d + W ), where gw is the weak coupling constant and θC is the Cabibbo angle (sin θC 0.22). The phase space terms can be estimated using the so called Sargent rule, originally established for the beta decay, taking into account the kinematic analogy of the ∝ 5 present decays with the beta case. Following this rule we have w E0 , where w is the transition rate and E0 is the energy available in the decay (= mn − m p − me, − for the beta decay n → p + e +¯νe). We also recall that in the Fermi theory the beta decay is only ‘kinematical’, that means that the energy dependence is only ρ ∝ 5 due to phase space. Therefore for the phase space we can write E0 . Under this assumption we have BR(D0 → K −e+ν ) cos2 θ m − m − m 5 e = C × D K e 20 × 0.32 6.4 ( 0 → π − +ν ) 2 − − BR D e e sin θC m D mπ me

Despite the crudeness of the estimate, this results differs from the experimental value by only 40%. Exercise 2.3.3 The Feynman diagrams are reported below

Exercise 2.3.4 The beta decay rate in the limit of the Sargent rule, i.e. assuming E mc2 and substituting E0 with Tmax( 0.782 MeV) is

G2 T 5 ω = F max . (2.1) 2π 37c6 30

The squared Fermi constant (divided by (c)3, as it is usually expressed) is 2 3 1 G ω 2π (c) 30 × 62 × 197 MeV fm × 30 − − F = = 886 s 4.710 21 MeV 4, ( )3 5 · 23 −1 × ( . )5 c cTmax 3 10 fm s 0 782 MeV 92 Appendix: Solutions of Exercises and Problems hence we have G F 6.910−11 MeV−2 = 6.910−5 GeV−2. (c)3

The value is different from the one reported in the literature (1.17 10−5 GeV−2) because of the spectrum integration inaccuracy implicit in the Sargent rule and other aspects of Fermi theory not included in Eq. (2.1), e.g. the V − A feature of weak interaction and the quark structure of the neutron. 35 →35 + − +¯ν Using the Sargent rule we have for the 16S 17 Cl e e decay ω[35S] Q[35S] 5 0.168 5 = = 0.00046, ω[n] Q[n] 0.782 and then 886 s τ[35S]= 1.9106 s 22 d. 0.00046 Both the parent and daughter nuclei are odd-A, hence the spin-parity is determined by the unpaired nucleon. The shell occupation of this nucleon is 35 : ( )2( )4( )2( )6( )2( )3 – 16S n 1s1/2 1p3/2 1p1/2 1d5/2 2s1/2 1d3/2 35 : ( )2( )4( )2( )6( )2( )1 – 17Cl p 1s1/2 1p3/2 1p1/2 1d5/2 2s1/2 1d3/2 = 35 35 = / , = (− )2 =+ The unpaired nucleons have l 2: both 16S and 17Cl have J 3 2 P 1 =⇒ J P = 3/2+.

Exercise 2.3.5

(1) Charged current νμ-interactions on nucleon valence quarks can be either

− νμ + d → μ + u or the ones associated to charm production

− νμ + d → μ + c.

In these processes, the leptonic vertex is the same whereas the hadronic one is gW cos θC in the former and gW sin θC in the latter case, where θC is the Cabibbo angle (sin θC 0.22). The fraction of charm events in CC interactions can be estimated as

σ(ν + → μ− + ) 2 θ μ d c = sin C = − − 2 2 σ(νμ + d → μ + u) + σ(νμ + d → μ + c) cos θC + sin θC

2 = sin θC 0.05 Appendix: Solutions of Exercises and Problems 93

(2) The probability for muon neutrinos to be detected as tau neutrinos is Pμτ whereas 1 − Pμτ is the probability to survive in the initial state. The signal-to-noise ratio is then

− N(ντ → τ ) Pμτ 1 0.015 1 r = × × 0.31. − 2 2 N(νμ + d → μ + c) 1 − Pμτ sin θC 0.985 0.22

− − − (3) The τ decay modes are of type τ → W + ντ . A few cases are given below − − − − • τ → μ +¯νμ + ντ [W → μ +¯νμ] − − − − • τ → e +¯νe + ντ [W → e +¯νe] − − − • τ → π + ντ [W → d +¯u] • etc. The Feynman graphs for these decays are shown below.

Exercise 2.3.6 Following the Fermi golden rule, the branching ratio is proportional to the absolute square of the transition amplitude times the phase space factor. In the decays of the text we have

(− → + − +¯ν ) |M(− → + − +¯ν )|2 ρ(− → + − +¯ν ) BR n e e n e e × n e e − − − − 2 − − BR( → + e +¯νe) |M( → + e +¯νe)| ρ( → + e +¯νe)

− For → n we have an effective coupling constant gw sin θC (dds → ddu involves − − s → u + W ), with θC the Cabibbo angle. Instead for → we have gw cos θC (dds → uds involves d → u + W −). The phase space factors (ρ) can be estimated using the Sargent rule. It is written ∝ 5 as w E0 , where w is the decay rate and E0 is the energy available in the decay 94 Appendix: Solutions of Exercises and Problems

− (= mn − m p − me, in the case of the neutron decay n → p + e +¯νe). In the Fermi theory of beta decay the transition rate is entirely due to kinematics. Therefore we can use the same expression for the phase space factor. So we have (− → + − +¯ν ) 2 θ − − 5 BR n e e sin C × m mn me − − 2 BR( → + e +¯νe) cos θC m − m − me and then − − 5 BR( → n + e +¯ν ) m − m − m 2 θ e × e tan C − − BR( → + e +¯νe) m − mn − me 10.2 1197 − 1116 5 × 0.056 0.57 1197 − 940

Hence sinθC 0.23, which is in good agreement with the known value (sinθC 0.22).

Exercise 2.3.7 + ¯ 0 + Considering the transition amplitudes, for D → K + e + νe we have c → s + + + + W and an effective coupling constant gW cos θC ;forμ → e + νe +¯νμ we have + + a pure leptonic vertex μ →¯νμ + W and thus only gW . Making use of the Sargent rule for the phase space factors we get

( + → ¯ 0 + + + ν ) |M( + → ¯ 0 + + + ν )|2 ρ( + → ¯ 0 + + + ν ) D K e e = D K e e × D K e e + + + + 2 + + (μ → e + νe +¯νμ) |M(μ → e + νe +¯νμ))| ρ(μ → e + νe +¯νμ) 5 5 + − − − − . 2 m D m K¯ 0 me 2 1870 498 0 5 5 = cos θC × 0.95 × 3.5 × 10 mμ − me 106 − 0.5

The experimental value is 1.5 × 105.

Exercise 2.3.8 − − + Pions produced in the atmospheric showers decay as π → μ +¯νμ and π → + − − + μ + νμ. The muons produced in this way decay as μ → e +¯νe + νμ and μ → + e + νe +¯νμ. All the charges have the same probability. The pions produced in the hadronic interactions with atmosphere nuclei have energies higher than the ones observed for the atmospheric neutrinos. Let us assume that the pion energy is at most 1 GeV. The pion mean free path is

pπ 1 8 −8 lπ = βγcτπ = cτπ < 310 × 2.610 m 55 m. mπ 0.140

Since their production height is around 10km, all the pions decay before reaching the ground, unless they interact with the atmosphere again. Under the same assumption, Appendix: Solutions of Exercises and Problems 95 the produced muons have a mean free path

pμ 1 5 −6 lμ = βγcτμ = cτμ < 310 × 2.210 km 6km mμ 0.106 and also muons preferentially decay. Counting all the types of neutrinos appearing in the decays we obtain a ﬂavor ratio

νμ +¯νμ 2 νe +¯νe

Exercise 2.3.9 − − − − The Feynman graphs for μ → e +¯νe + νμ and τ → e +¯νe + ντ are identical apart the masses involved. Recalling the Sargent rule, one gets: − − 5 5 (τ → e +¯ν + ντ ) mτ − m mτ = e = e . × 6 R − − 1 32 10 (μ → e +¯νe + νμ) mμ − me mμ

− − Denoting with B(τ → e +¯νe + ντ ) the branching ratio of this mode, the tau mean lifetime is then

τ . × −6 μ − − 2 2 10 −13 ττ = × B(τ → e +¯ν + ντ ) × 0.18 3 × 10 s R e 1.32 × 106

A.3 Solutions of Experiments and Detection Methods (Chapter 3)

3.1 Kinematics

Exercise 3.1.1 Here we write a few kinematical relations useful for the solution: = 2 + 2 = 2 + 2 π pπ mπ pπ mπ 20 GeV ∗ = ∗ + ∗ = 2 + 2 + . E π p mπ m p 2m p π 6 199 GeV βCM =|pπ |/(π + p) = pπ /(π + m p) 0.955165 γ = ( + )/ ∗ . CM π √p E 3 3775 [ ∗2−( + )2][ ∗2−( − )2] ∗ =| ∗|= E m m K E m m K . / p p 2E∗ 2 965 GeV c 1. Neglecting the thickness of the target, detectable tracks are produced by ionising particles emitted between 0◦ and 90◦ in the Laboratory system (LS). To answer the ﬁrst question we have to establish if +’s produced in the experiment do exhibit a maximum angle. We have ∗ ∗2 2 = p + m 3.194 GeV 96 Appendix: Solutions of Exercises and Problems

∗ ∗ ∗ β = p / 0.9281

∗ + β <βCM is the condition to have such limiting angle and hence all ’s can be detected.5 2. Assuming that all +’s decay within 3 mean lifetimes, the maximum distance for the decay point is

p p D = 3 · cτ · βγ = 3 · cτ · 6.05 × cm, m GeV/c along the direction p/p. From this expression we desume that the minimum length for the tracker corresponds to the maximum longitudinal momentum (p)L . ∗ ∗ This occurs for (p )L = p . Hence we have

( )max = γ ( ∗ + β · ∗ ) . / =⇒ p L CM p CM 20 3GeV c

L = 6.05 × 20.3cm 122.8cm

3. As in the previous case, the minimum radius corresponds to the maximum (p)T , ∗ ∗ that is for (p )T = p

max (p) R = 6.05 × T = 6.05 × 2.965 cm 17.9cm GeV/c

4. To establish if there is a maximum angle for K +, we calculate its velocity in the CMS ∗ = ∗2 + 2 . K p m K 3 006 GeV

β∗ = ∗/∗ . K p K 0 9864 β∗ >β We have K CM, so there is no limiting angle. Hence kaons can escape from the tracker.

5It can be useful, though not necessary, to calculate the maximum angle and its corresponding angle in the CMS. They are θ = β∗ /(γ β2 − β∗ 2) . =⇒ θ . ◦ tan max CM CM 1 217 max 50 6 ∗ ∗ ∗ ◦ cos θ (θmax) =−β /βCM −0.9717 =⇒ θ (θmax) 166 At the maximum angle the longitudinal, transverse and total +-momenta are respectively

∗ ∗ ∗ (p)L = γCM (p · cos θ (θmax) + βCM · ) 0.573 GeV/c ∗ ∗ ∗ (p)T = (p)T = p · sin θ (θmax) 0.700 GeV/c p 0.905 GeV/c. . Appendix: Solutions of Exercises and Problems 97

5. The detectable kaons are those produced in the forward direction in the LS (0◦ < θ<90◦). The CMS angle corresponding to θ = 90◦ can be obtained from the Lorentz ( ) = γ [( ∗ ) + β · transformation of the kaon longitudinal momentum pK L CM pK L CM ∗ ] ( ) = . K by setting pK L 0 Hence we have ( ∗ ) = ∗ θ ∗( ◦) =−β ∗ =⇒ pK L p cos 90 CM K ∗ β θ ∗( ◦) =−β · K =− CM −. , cos 90 CM ∗ β∗ 0 9683 p K corresponding to angle of about 165.5◦. In the CMS frame the kaon angular distribution is isotropic so it is given by dN/d = 1/4π (normalized to unity). The fraction of detectable kaons is then 1 2π 1 1 − cos θ ∗(90◦) 1.9683 r = dφ d cos θ ∗ = 98.4% 4π 0 θ ∗(90◦) 2 2

Exercise 3.1.2 β ≥ β∗ (1) To have a limiting production angle, particle 1 must fulﬁll the condition CM 1 . The maximum limiting angle, corresponding to 90◦, is obtained for the equality in the previous relation. The CMS energy is the mass of the resonance, E∗ = M. Hence we have

∗ ∗ 2 2 2 2 p =|p |= [M − (m1 + m2) ][M − (m1 − m2) ] / 2M

∗ = ( 2 + 2 − 2)/ 1 M m1 m2 2M and then ∗ [ 2 − ( + )2][ 2 − ( − )2] β∗ = p = M m1 m2 M m1 m2 1 ∗ 2 + 2 − 2 1 M m1 m2

Since m2 is negligible with respect to m1, we get

M2 − m2 2.582 − 1 β∗ = 1 = 0.7388 1 2 + 2 . 2 + M m1 2 58 1 β = β∗ We can get the pion beam energy Eπ , solving the equation CM 1

| | E2 − m2 β = pπ = π π = β∗. CM 1 Eπ + m p Eπ + m p

Solving it in Eπ we have 98 Appendix: Solutions of Exercises and Problems β∗2 + (β∗2 )2 + ( − β∗2)( 2 + β∗2 2 ) 1 m p 1 m p 1 1 mπ 1 m p Eπ = 2.65 GeV. − β∗2 1 1

◦ For higher Eπ values particle 1 is produced up to angles less than 90 . (2) Considering the decay (2420) → + K , for a ﬁxed CMS angle θ ∗ = 120◦ and βCM = 0.7388, we have

[M2 − (m + m )2][M2 − (m − m )2] p∗ =|p∗|= K K 0.833 GeV/c 2M ∗ ∗2 2 = p + m 1.452 GeV

∗ ∗ ∗ (p)L = γCM (p cos θ + βCM · ) 0.974 GeV/c

∗ ∗ (p)T = p sin θ 0.721 GeV/c.

From the last expressions we can get the momentum and angle of the in the Laboratory system = ( )2 + ( )2 . / p p L p T 1 21 GeV c

(p)L ◦ θ = arccos 36.5 p

(3) The -decay mean pathlength is cτ · βγ = cτ · p/m. The length of the detector is determined by the ’s decaying in the forward direction, for which the momentum is maximum. This is

max ∗ ∗ p = γCM (p + βCM · ) 2.83 GeV/c

The designed length corresponds to the requirement that 99% of the decay points are contained in the detector. This occurs for a proper time T so that we have T 1 ∞ 1 t T 0.99 = N(t)dt = 1 − exp − dt = 1 − exp − 0 τ N0 T τ τ τ

=⇒ T =−ln(0.01) · τ 4.6 · τ

Hence the length of the detector must be

max max p p L = cT· 4.6 c τ · m m Appendix: Solutions of Exercises and Problems 99

Solving this equation in τ, we ﬁnally get . × . Lm 0 26 1 189 −10 τ = 0.79 × 10 s max 8 4.6 cp 4.6 × 3 · 10 × 2.83

Exercise 3.1.3

1. The minimum energy for a reaction is its threshold energy (Eth). It corresponds to the production of the ﬁnal particles at rest in the CMS. Equating the 4-momentum invariants in the LS for the initial state and in the CMS for the ﬁnal state, we have

2 + 2 + = ( + )2 Mπ Mp 2Eth Mp M MK and then ( + )2 − 2 − 2 M MK Mπ Mp Eth = 0.91 GeV 2Mp

∗ ∗ 2. A maximum production angle is possible provided that β <βCM, where β is the CMS velocity of the particle and βCM is the velocity of the CMS with respect to the LS. For Eπ = 2GeV,wehave

pπ βCM = 0.68 Eπ + Mp

To get the velocity in the CMS, we ﬁrst calculate the total CMS energy (Pπ and Pp are the 4-momenta of the pion and proton respectively) ∗ = ( + )2 = 2 + 2 + . E Pπ Pp Mp Mπ 2Eπ Mp 2 16 GeV

The momentum in the CMS is

∗ ∗ [E 2 − (M + M )2][E 2 − (M − M )2] p∗ =|p∗|= K K 0.69 GeV/c 2 E∗ Hence we have for the -velocity in the CMS

p∗ β∗ = 0.52. ∗2 2 p + M

∗ The condition β <βCM is fulﬁlled so that there is a maximum production angle for the ’s. This angle turns out to be ⎧⎡ ⎤ ⎫ −1 ⎨ β 2 ⎬ θ = ⎣γ CM − ⎦ . ◦ max arctan ⎩ CM β∗ 1 ⎭ 0 73 rad 42 100 Appendix: Solutions of Exercises and Problems

Exercise 3.1.4

The invariant mass of the two-pion system is the mass MX of the observed neutral particle

2 = ( + + − )2 = 2 + + − − + − θ. MX Pπ Pπ 2Mπ 2Eπ Eπ 2pπ pπ cos

The minimum opening angle corresponds to the case in which the two pions have the same energy Eπ + = Eπ − (= E X /2). Imposing this condition and having in mind that Eπ Mπ we get = 2 2 θ/ + 2 . / 2 MX E X sin 2 2 Mπ 0 495 GeV c

Exercise 3.1.5

Denoting by Pi the 4-momentum of electron i, the total CMS energy is written as

∗ 2 2 2 E = (P1 + P2) = (E1 + E2) − ( p1 + p2) = 2 + 2 + − 2 − 2 + = 2 + + E1 E2 2E1 E2 p1 p2 2p1 p2 2m 2E1 E2 2p1 p2 4E1 E2 where, in the last step, we have neglected the electron masses with respect to their energies. Therefore we have E∗ = 15.5GeV. In the CMS the two electron momenta are opposite. Neglecting the masses we have E∗ p∗ = 7.74 GeV/c. 2 The CMS velocity (in c units) in the LS is given by E2 − m2 − E2 − m2 | p1 + p2| 1 2 E1 − E2 βCM = = 0.4 (3.1) E1 + E2 E1 + E2 E1 + E2 and the Lorentz factor is

γ = ( − β2 )−1/2 . . CM 1 CM 1 1

If E1 = E2 and p1 =−p2, from Eq. (3.1) we get βCM = 0. Hence the center-of- momentum and laboratory systems are coincident.

Exercise 3.1.6 The minimum electron energy is its rest mass ( 0.511 MeV), corresponding to the emission of an electron at rest. To evaluate the maximum energy in a three-body decay M → m1 + m2 + m3,it is convenient to re-write it as a two-body decay M → M12 + m3, where M12 is the Appendix: Solutions of Exercises and Problems 101 invariant mass of particles 1 and 2. It then turns out that the maximum energy for 3 is obtained when M12 is minimum, that is when it is equal to the m1 + m2. Hence we have M2 + m2 − (m + m )2 (E ) = 3 1 2 . 3 max 2M

In our case M = M0 , m3 = Me− , m1 = M+ and m2 = Mν = 0 and we have

13152 + 0.5112 − 11892 (E − ) 120 MeV e max 2 × 1315

Exercise 3.1.7 The minimum opening angle for a decay into two equal (ultra-relativistic) particles is obtained for E D Eπ + = Eπ − = . 2 Hence for the minimum opening angle between the pions we have ⎛ ⎞ ⎛ ⎞ M2 − 2M2 M2 − 2M2 ⎝ D π ⎠ ⎝ D π ⎠ θmin = arcsin = arcsin 2 (3.2) Eπ E D

¯ To get the D0 energy, we make use of the knowledge that this particle is produced at the maximum angle θmax. The corresponding angle in the CMS is given by the equation β∗ ¯∗ ∗ D cos θ = cos θ (θmax) =− , βCM

β∗ ¯0 β where D is the D velocity in the CMS and CM is the CMS velocity in the LS. ∗ ∗ ¯ 0 Denoting by E D and p the energy and momentum of the D -particle in the CMS, using the Lorentz transformation for the energy, we get

= γ ( ∗ + β ∗ θ¯∗) = γ ( ∗ − β∗ ∗). E D CM E D CM p cos CM E D D p (3.3)

The total energy in the CMS is ∗ = ( + )2 = + 2 + 2 = . . E Pπ Pp 2Eπ Mp Mπ Mp 6 21 GeV

Hence the D¯ 0-momentum in the CMS is

∗2 ∗2 [E − (M − M )2][E − (M + M )2] p∗ = D D 2.22 GeV/c 2E∗ 102 Appendix: Solutions of Exercises and Problems ∗2 + 2 . and the corresponding energy is p MD 2 90 GeV. For the quantities appearing in Eq. (3.3) we get ∗ + β∗ = p = . γ = Eπ Mp = . , D 0 767 CM ∗ 3 37 ∗2 + 2 E p MD and hence we have E D = 4.0 GeV. Using (3.2), we ﬁnally get for the minimum opening angle # √ $ 1.862 − 2 × 0.1402 θ = arcsin 2 × 1.18 rad 67.6◦ min 4

Exercise 3.1.8 Equating the 4-momentum invariants in the LS for the initial state and in the CMS for the ﬁnal state, we have at the threshold

2 + 2 + − 2 − 2 − · = ( + )2 E p Eγ 2E p Eγ pp pγ 2 pp pγ Mp Mπ

In the UHE regime we assume E p ≈ pp and then we get

( − θ) = ( + )2 − 2 2E p Eγ 1 cos Mp Mπ Mp

The threshold energy as a function of the scattering angle is then

( + )2 − 2 Mp Mπ Mp Eth(θ) = . 2Eγ (1 − cos θ)

The minimum value is obtained in the case of head-on scattering, θ = π

2 2 (M + Mπ ) − M min = p p . × 19 . Eth 6 8 10 eV 4Eγ

Exercise 3.1.9 Using the relativistic invariants we have

( + )2 = ( + )2 − ( + )2 = 2 + − · . Mp Mn Ed Eγ pd pγ Md 2Eγ Ed 2 pd pγ =− From the momentum conservation we can write pd pγ and then

( + )2 = 2 + ( − ) 2 + ( + )2, Mp Mn Md 2Eγ Ed pd Md 2Eγ Md Md Eγ Appendix: Solutions of Exercises and Problems 103 where, in the last two steps, we have considered that both the deuteron recoil momentum the photon energy (both O(MeV)) are negligible with respect to the deuteron mass. Hence we can write

Md Mp + Mn − Eγ .

Assuming that the proton and neutron masses have negligible errors, it follows that Md = Eγ , and ﬁnally we get

2 Md = 1875.607 ± 0.005 MeV/c

Exercise 3.1.10 1. The total CMS energy is ∗ = 2 + . E 2Mp 2Mp E p¯ 2 08 GeV = 2 + 2 = were we used E p¯ pp¯ Mp 1.37 GeV. The kaons in ﬁnal state are produced ◦ ∗ = ∗/ back-to-back at 90 in the CMS. Their energies are EK E 2 and the momenta are ∗ = ( ∗/ )2 − 2 . / . pK E 2 MK 0 92 GeV c

∗ = ∗ ∗ = Therefore in the CMS pT pK and pL 0. Using the Lorentz transformation to theLSweget ( ) = ∗ . / pK T pT 0 92 GeV c p ¯ (p ) = γ [p∗ + β E∗ ]=0 + p (p∗ )2 + M2 0.50 GeV/c K L CM L CM K E∗ K K The kaon energy in the LS is then = [( )2 + ( )2 ]+ 2 . EK pK T pK L MK 1 16 GeV and the production angle is (pK )T ◦ θK = arctan 61.5 (pK )L

The kaons we are detecting have βγ = pK /MK = 2.1 and we can then assume that dE = MeV their energy loss in the gas is dx ion 2 gcm−2 . The number of electron-ion pairs in each detector turns out to be 6 1 dE 210 −3 n = ρ d p c × 210 × 10 × 0.20 × 0.30 160. I dx ion 15 104 Appendix: Solutions of Exercises and Problems

Exercise 3.1.11 The mass of the particle is obtained from the invariant mass of the two muons. Their energies in the LS are = 2 + 2 2 + 2 E1 p1 mμ 45 106 115 MeV = 2 + 2 = . E2 p2 mμ p2 30 GeV

The square of the total 4-momentum is the invariant mass of the system. The total energy and momentum are

Et = E1 + E2 0.115 + 30 30.12 GeV

| |=| + |=| |−| |= − . = . / pt p1 p2 p2 p1 30 0 045 29 96 GeV c

Hence for the mass we have 2 2 2 2 2 M = Et − pt = 30.12 − 29.96 = 3.10 GeV/c .

The particle is the J/ψ-meson. Exercise 3.1.12 At the threshold we have

2 + 2 + − 2 − 2 − · = ( + )2, E p Eγ 2E p Eγ pp pγ 2 pp pγ Mp 2me and for E p pp:

( − θ) = ( + )2 − 2 . 2E p Eγ 1 cos Mp 2me Mp 4Mpme θ Substituting EγCMB to Eγ , the threshold energy as a function of turns out to be

2M m E (θ) p e . th ( − θ) EγCMB 1 cos

The minimum value is obtained for θ = π (head on scattering) and is

M m min = p e . × 18 . Eth 0 5 10 eV EγCMB

Exercise 3.1.13

(a) Denoting by s the square of the total energy at LHC and with ELab the energy in ﬁxed target pp interactions, we require Appendix: Solutions of Exercises and Problems 105

s = 2 mELab where m is the proton mass. Here we have assumed that protons are ultra-relativistic. Hence we get

s (13 1012)2 eV2 E = 9 × 1016 eV Lab 2 m 2 × 0.94 109 eV

(b) Denoting by v the insect velocity we have

1 Mv2 = E , 2 Lab since in the ultra-relativistic limit the proton kinetic energy is almost equal to its total energy. Then we get % 2 E 2 × 91016 × 1.610−19 J v = Lab 11 m/s 39 km/h M 0.25 10−3 kg

Exercise 3.1.14 (a) Consider the Lorentz transformation between the reference systems K’ and K. Denoting by β the velocity of K’ with respect to K,wehave

E = γ(E + βp), p = γ(p + β E ).

Hence we get

E ± p = γ(1 ± β)(E ± p),

+ + β E + p E p = 1 ×

E − p 1 − β E − p

Using the deﬁnition of rapidity we ﬁnally obtain 1 + β y = y + ln 1 − β

(b) The maximum (minimum) rapidity is obtained for the elastic scattering, pp → ◦ ◦ pp,atθ = 0 (θ = 180 ). The maximum value is then for p = p and p equal to the beam momentum (6.5 TeV)

1 E + p 1 (E + p)2 E + p 6500 + 6500 y = ln = ln = ln = ln 9.5 max 2 E − p 2 E2 − p2 m 0.94 and ymin =−ymax. 106 Appendix: Solutions of Exercises and Problems

1 E(1 + cos θ) 1 cos2 θ/2 θ y ln = ln =−ln tan = η 2 E(1 − cos θ) 2 sin2 θ/2 2

(d) At 90◦ rapidity and pseudorapidity are identical: y = η = 0. At 1◦ we have √ 1 p2 + m2 + p cos 1◦ 1 65002 + 0.942 + 6500 cos 1◦ y = ln = ln √ 4.74123 2 p2 + m2 − p cos 1◦ 2 65002 + 0.942 − 6500 cos 1◦

η =−ln tan 0.5◦ 4.74134.

Therefore the difference is of the order of 1 over 105.

Exercise 3.1.15 For a particle moving along the x direction and emitting a decay particle at angle θ after a (proper) time t,wehave p = ctβγ sin θ = ct sin θ m Denoting CMS quantities with * and with no index the ones in the LS, we obtain from the Lorentz transformations

= ∗,= γ(∗ + β ∗) py py px

p sin θ = p∗ sin θ ∗,= γ∗(1 + ββ∗ cos θ ∗), where β∗ is the velocity of the emitted particle in the CMS (= p∗/∗). From their ratio we get β∗ sin θ ∗ βγ sin θ = 1 + ββ∗ cos θ ∗

In the ultra-relativistic limit for both particles (β → 1, β∗ → 1) we have

sin θ ∗ θ ∗ βγ sin θ → = tan 1 + cos θ ∗ 2 and ﬁnally obtain θ ∗ → ct tan 2 which proves that the impact parameter is independent from the particle momentum. The mean value of the impact parameter for t = τ is Appendix: Solutions of Exercises and Problems 107 π θ ∗ sin θ ∗dθ ∗ (x − sin x)|π π =cτ tan = cτ 0 = cτ 0 2 2 2 2

For the D+ decay we get

1.57 × 3108 m/s × 1.04 10−12 s 490 µm

For a CMS angle θ ∗ = 90◦ we obtain

3108 m/s × 1.04 10−12 s × tan 45◦ 312 µm

Exercise 3.1.16 The mass of the parent particle is the invariant mass of the two muons. We have

2 = 2 = ( + )2 = 2 + 2 + − | || | θ, M P p1 p2 m1 m2 2E1 E2 2 p1 p2 cos where pi and pi are respectively the 4-momentum and the momentum of particle i (=1, 2). In the ultra-relativistic limit, holding for both muons, we get θ M2 = m2 + m2 + 4E E sin2 . (3.4) 1 2 1 2 2

Substituting m1 = m2 = mμ and considering that the muon mass is negligible with respect to the energies of both muons, we get % θ θ 2 2 M = 2mμ + 4E E sin 2 E E sin 1 2 2 1 2 2

√ 42◦ 2 7.4 × 2.6sin 3.1GeV/c2 2 The mass value corresponds to the one of J/ψ. The momentum of the particle can be obtained from the Carnot theorem = 2 + 2 + θ = . 2 + . 2 + × . × . ◦ . / p p1 p2 2p1 p2 cos 7 4 2 6 2 7 4 2 6 cos 42 9 5GeV c and its energy is then E = p2 + M2 10 GeV. Substituting E2 = E − E1 in (3.4), we can express the opening angle as a function of E and E1: θ M2 − m2 − m2 sin = 1 2 2 4E1(E − E1)

This expression is minimum for E1 = E/2 108 Appendix: Solutions of Exercises and Problems 2 − 2 − 2 θ M m1 m2 M 3.1 sin = = 0.31 2 min E E 10 which corresponds to an opening angle of 36◦. The energy of the muons is then 10/2 =5GeV. Exercise 3.1.17

(a) The CMS energy of a particle emitted in a two-bosy decay is (M = mπ )

2 + 2 − 2 2 − 2 2 2 ∗ mπ mν mμ mπ mμ 0.140 − 0.106 ν = = 30 MeV 2mπ 2mπ 2 × 0.140

(b) For a 200 GeV pion the Lorentz factor and the velocity are γ = p/mπ 200/0.140 1429 and β ≈ 1 respectively. Transforming the neutrino energy to the LS we get ∗ ∗ ∗ ∗ ∗ Eν = γ(ν + βpν cos θ ) = γν (1 + β cos θ ) where θ ∗ is the neutrino emission angle in the rest frame. The maximum energy is obtained for θ ∗ = 0 and is

∗ ∗ Eν (max) = γν (1 + β) 2γν 2 × 1429 × 30 MeV 85.7GeV

∗ ◦ ∗ ◦ ∗ Eν (max) Eν (θ = 90 ) = γν (1 + β cos 90 ) = γν = 42.9GeV. 2 Therefore forward emitted neutrinos have energies larger than this value. (d) Using the relationship between angles under a Lorentz transformation we have

sin θ ∗ tan θ = γ(cos θ ∗ + β)

For neutrinos in the forward hemisphere in the CMS the maximum angle corresponds to θ ∗ = 90◦ θ = 1 1 . tan max γβ γ 0 00070 which is an angle of 0.04◦. Exercise 3.1.18 If p and m are respectively the momentum and mass of the neutron, the distance it has to cover in a mean lifetime is

p cT / L = γβ c τ = 1 2 , m ln 2 Appendix: Solutions of Exercises and Problems 109

Fig. 3.1 Feynman diagram for γ + γ → e+ + e−

where L( 5000 × 365 × 24 × 60 c × minutes) is the distance of the source. Hence we get

L × ln 2 2.6109 min × 0.693 p × m × m 1.8108 × 0.940 GeV 1.71017 eV/c T1/2 10 min

Neutrons with such momentum are ultra-relativistic and thus also their energy has the same value.

Exercise 3.1.19 (a) The lowest order Feynman diagram is shown in Fig. 3.1. The amplitude is proportional to α and the cross section to α2. (b) Let us write (E, K) the 4-momentum of the photon from the source and (, k) the one of the CMB photon, in the LAB system. K and k are opposite (head-on) and, considering also that they are massless, their sum is E − . At the threshold we have:

2 2 2 2 2 2 P = (E + ) − (K + k) = (E + ) − (E − ) = (2me)

= 2, 4E 4me and therefore the minimum photon energy is

m2 (0.511 × 106)2 E = e 2.6 × 1014 eV 10−3 (c) Denoting by M the invariant mass, at the threshold the Lorentz factor of the CMS system is E + E + γ = = 2.5 × 108 M 2me 110 Appendix: Solutions of Exercises and Problems

3.2 Interaction of Radiation with Matter

Exercise 3.2.1 The number of photons as a function of the matter thickness x is

−μx N(x) = N0 e where μ = 0.04 cm2/g for lead and x is expressed in g/cm2. To halve the number of photons we require

N0 −μx / ln 2 2 = N e 1 2 =⇒ x / = 17.33 g/cm . 2 0 1 2 μ

ρ = . / 3 = x1/2 = . Using 11 3g cm ,wehavel1/2 ρ 1 53 cm. For a 5% photon survival we have 1 . N = N e−μx =⇒ x =− ( . ) / 2, 0 05 0 0 μ ln 0 05 75 g cm and then l5% = 6.63 cm. Exercise 3.2.2

In the high energy limit (Eγ me) the absorption coefﬁcient for pair production is 7 μ = X −1 1.4cm−1 9 0

The cross section can be obtained from the absorption coefﬁcient μ using the relationship ρN μ = nσ, where n = A . A

3 23 −1 Using A = 207, ρ = 11.3g/cm and the Avogadro number NA = 6.02 · 10 mole we get A μ σ = 4.2 · 10−23 cm2 = 42 b NA ρ

Exercise 3.2.3 Neglecting the momentum loss in the slab (see below), the radius of curvature of the muon is p R = 33.3m, 0.3 B where, in this equation, B, R and p are given in Tesla, meter and GeV/c respectively. The muon deﬂection angle θ is equal to the angle of the radius at the exit with respect to the slab. For small angles the circular segment can be approximated to the slab thickness (see ﬁgure) and then we can write Appendix: Solutions of Exercises and Problems 111 L θ arcsin 0.015 rad 0.86◦ R

Considering its initial energy the muon energy loss rate corresponds to the one of − dE . 2/ a minimum ionising particle, dx 1 4 MeV cm g. The energy loss is then dE E = − × ρl 550 MeV. dx

= θ E = p where l R L 50 cm. Since E p , it follows that the muon momentum after the slab is p = p − p p − E 19.5GeV/c.

The multiple scattering dispersion in the plane of the ﬁgure is given by E l θ 2 =√ s . ◦ s 4mrad 0 23 2p¯β X0 √ where Es 20 MeV is the multiple√ scattering constant and p¯= pp 19.75 GeV/c (see Exercise 3.2.8). The factor 2 at denominator converts the spatial dispersion θ 2 = θ 2 / angle to the plane angle ( proj space 2).

Exercise 3.2.4 2 The Compton scattering cross section in the low energy limit, Eγ mec ,isgiven by the Thomson cross section 8 σ = πr 2, 3 0

2 2 where r0 is classical electron radius [= e /(4π0mc ) 2.8 fm]. Hence we have 112 Appendix: Solutions of Exercises and Problems

1 1 λ = = 4.52 cm σn 8 π 2 Z ρ 3 r0 NA A

Exercise 3.2.5 Using the equation p[GeV/c]=0.3 × B[T]×R[m] and writing the sagitta as s L2/(8R), valid for R s, the electron momentum is

L2 0.032 p = 0.3 B 0.3 × 0.1 1.7MeV/c. 8s 8 × 0.002

The kinetic energy of the electron is T = p2 + m2 − m 1.3MeV. The 4-momentum conservation in the Compton scattering can be written as Eγ + m = Eγ + E k = k + p, being (Eγ , k) and (m, 0) the initial 4-momenta of the photon and electron, and (Eγ , k ) and (E, p) the ﬁnal ones. Squaring k − p = k and solving in Eγ , we get for the initial photon energy6 p2 − T 2 Eγ = 1.6MeV. 2(pcosφ − T )

The scattered photon energy is

Eγ = Eγ − T 0.3MeV

Exercise 3.2.6 The mean number of pairs created by a single pion is − dE ρ d n = dx ion . I

In this exercise, as in many others in this book, the value of the ionization loss rate is not given for the speciﬁc case (particle, material, etc.). Most of the cases refer to relativistic singly charged particles. To help making a correct estimate one should have in mind the main features of (−dE/dx)ion that can be easily deduced from a ﬁgure of this function, e.g., as reported in the PDG Review of Particle Physics [1]. These features can be summarized as follows:

• the minimum of (−dE/dx)ion is at βγ ≈ 3. The differences in the minimum ionization loss rate among the different materials is modest, because it is mainly determined by the ratio Z/A: they change from ≈1.2MeVg−1 cm2 forPbupto

6It should not surprise that the scattered electron has a momentum larger than the initial photon energy. In fact the photon is scattered backward at an angle of about 108◦. Appendix: Solutions of Exercises and Problems 113

≈2MeVg−1 cm2 for He. The only exception is hydrogen, whose Z/A is 2 and about twice w.r.t all the other elements, which has a minimum ionization energy loss rate of ≈4MeVg−1 cm2. • In the relativistic and ultra-relativistic regimes, the increase of (−dE/dx)ion with βγ is very small and in some case negligible. To have a reference number, there is a factor of about 1.5 with respect to the minimum ionization in the βγ range from 3 to 10,000. • The previous consideration is actually true only for solid and liquid materials. In these materials the energy loss is modiﬁed at increasing βγ by the so-called “density effect”. Instead for gases this effect is negligible and the increase of (−dE/dx)ion with βγ is somewhat larger (in the βγ range from 3 to 10,000 a factor about 2).

20 GeV pions have βγ 140 and the medium in the counter is a gas (whose composition is not given). Taking into account the fact that the typical gases used in ionization counters have a minimum ionization of 1.5 ÷ 2MeVg−1 cm2 and a contribution due to the relativistic increase in gases, we can assume an energy loss rate dE MeV cm2 − 2 . dx ion g

Using d =1cm, ρ = 1.8 × 10−3 g/cm3 and I =15 eV, we get n = 240 pairs. Hence we have for the current

Iout = I0 × n = 2.4mA

Exercise 3.2.7 The velocity for protons and pions are p p βp = 0.983 βπ = 0.9996 2 2 2 + 2 p + Mπ p Mp

Hence they are related as

1 βπ >βp > = 0.952 n1

So the ﬁrst Cherenkov detector is sensitive to both particle types. To get the beam separation we then require a refractive index in the second detector allowing the detection of the faster particle only

1 βπ > >βp n2 from which we get 1.0004 < n2 < 1.017. 114 Appendix: Solutions of Exercises and Problems

Exercise 3.2.8 500 MeV/c muons have βγ 4.7 and so they are close to the ionization minimum. For the copper slab we can assume dE MeV − = 1.4 . −2 dx ion gcm

The thickness to stop the muon beam is the range for these muons. A simple estimate can be done assuming that the energy loss is constant along the particle trajectory 1 T dT R = ρ 0 (−dE/dx)ion

1 T p2 + M2 − M 405 MeV = 32 cm, ρ (−dE/dx)ion ρ(−dE/dx)ion 9 × 1.4MeV/cm where T and p are the initial muon kinetic energy and momentum respectively, and M is their mass. A better value for the range can be obtained from the graph R/M versus βγ shown in the ﬁgure below, taken from the PDG Review of Particle Physics [1]. From this ﬁgure we deduce for an element (Fe) close to the copper R/M 2300 g cm−2/GeV. Substituting to M the muon mass we get R 27 cm. Appendix: Solutions of Exercises and Problems 115

The kinetic energy lost in a 10cm slab is dE T = − ρ d = 126 MeV. dx ion

Hence the mean energy of the muons after the slab crossing is

T = T − T 279 MeV.

The multiple scattering angle at the exit has to be calculated taking into account the ionization energy loss in the slab, because this loss is not negligible T/T 126/405 31%. The calculation has to be done as follows E 2 dx E 2 dpβ dθ 2 = s = s . pβ X0 pβ X0 (−dpβ/dx)ion

2 2 where X0, converted to g/cm ,is1.4 × 9 = 12.6g/cm and (−dpβ/dx)ion,thepβ loss rate, can be obtained from the ionization energy loss rate as β β −dp = −dE dp . dx ion dx ion dT

We can write p2 (T + M)2 − M2 M pβ = = = T 1 + (3.5) E T + M T + M hence we get & ' dpβ dE M 2 dE − = − + = − [ + (T )] 1 + 1 dx ion dx ion T M dx ion

The function (T ) is ∼4% for the entrance energy and ∼7.5% for the exit energy. For an estimate of the scattering angle (within an accuracy of less than 10%) we can neglect such function in the previous expression and calculate the r.m.s. scattering angle as

( β) ( β) p f E2 p f dpβ θ 2 = θ 2 = dθ 2 = s = s (− / ) ( β)2 (pβ)i X0 dE dx ion (pβ)i p

E2 (pβ) − (pβ) = s f i , (3.6) X0 (−dE/dx)ion (pβ)f (pβ)i 116 Appendix: Solutions of Exercises and Problems where (pβ)i 489 MeV/c and (pβ)f 356 MeV/c are the pβ value corresponding to the entrance and exit of the muons. In the previous expression we have assumed constant the energy loss rate within the integration range. Hence we obtain

202 489 − 356 θ 2 = 0.017 s 12.6 · 1.4 489 · 356 and ◦ θs 0.132 rad 7.5 .

Equation (3.6) allows to get a simple rule to calculate the multiple scattering angle to be used in case of sizeable energy loss. In fact, considering that we have (pβ)f − (pβ)i ≈ (−dE/dx)ion × d, the r.m.s. scattering angle can be written as in the case of pβ constant Es d θs = [pβ] X0 replacing pβ with the geometric mean [pβ]= (pβ)f (pβ)i 417 MeV/c.

Exercise 3.2.9 The energy of the photons which are incident on the silver foil is

hc 6.28 × 197 106 × 10−6 eV nm Eγ = 6.2eV. λ 200 nm

To have the photoelectric effect the photon energy must fulﬁll the condition Eγ > W. With W = 4.73 eV the photoelectric process is allowed. The electron kinetic energy is EK = Eγ − W = 1.47 eV.

Exercise 3.2.10 According to the Heitler toy model, the depth T at which the shower reaches the T maximum development is given by the equation 2 = E0/Ecrit. Hence we have

log (E /E ) log (100 GeV/80 MeV) T = 10 0 crit = 10 10.3 log10 2 log10 2 where T is given in units of radiation lengths. Hence the air thickness is

2 2 Xmax = T × X0 = 10.3 × 37 g/cm 380 g/cm

Exercise 3.2.11 + − The photons emitted in the e e annihilation at rest have energy Eγ = M/2 = me, being M = 2me and me the electron mass. In the Compton process the scattered photons have energy Appendix: Solutions of Exercises and Problems 117

Eγ Eγ = 1 + Eγ /me(1 − cos θ) from which it follows that the extremal electron kinetic energies are for cos θ = 1 and cos θ =−1

(Te)min = Eγ − (Eγ )max = Eγ − Eγ = 0

Eγ me 2 (Te)max = Eγ − (Eγ )min = Eγ − = me 0.34 MeV 2Eγ + me 3

Exercise 3.2.12 Assuming an energy loss rate of 2 MeV/(g · cm−2), the minimum kinetic energy for a vertical muon to reach ground7 is dE MeV T − × x 2 × 1030 g · cm−2 2.1GeV min · −2 dx ion g cm

The number of ionized electrons are

T 2.1 × 109 eV N = min 108. e I 10 eV

Exercise 3.2.13 We notice ﬁrst that the plate thickness is much smaller than a radiation length so that we can neglect the electron energy loss. In this condition the r.m.s. scattering angle is simply % E x 2 s θs = θ E0 X0 where E0 is the electron energy, Es 20 MeV and x = X0/20. Considering the electron bremsstrahlung the dispersion angle is about

me θb E0 where me is the electron mass. We have % 20 1 0.5 θ = 4.5mrad θ = 0.5mrad s 1000 20 b 1000 and then the angular distribution is dominated by the multiple scattering.

7The muon mean decay pathlength is cτ = 660 m. This value does not contrast with the total pathlength (10 ÷ 15km) from the muon production to the sea level. At the minimum energy the mean decay pathlength in the Earth reference system is Lμ = βγcτ Emin/mμ × cτ 2100/106 × 660 m 13 km. This means that most of the muons above the minimum energy reach the ground. 118 Appendix: Solutions of Exercises and Problems

Exercise 3.2.14 For muons (E = 3 GeV) in copper we can assume (see Exercise 3.2.6) a ionization energy loss rate dE MeV − = 1.8 . −2 dx ion gcm

In a d = 10cm thick slab muons lose an energy dE E = − ρd = 0.16 GeV, dx ion and so we have E E. The lateral beam broadening can be calculated assuming that the muon energy is unaffected by the slab crossing and we can write xS xS x2 E 2 x3 E 2 (ρr)2 = x2dθ 2 = dx s S s 0 0 X0 pβ 3X0 pβ

2 where xS = dρ 90 g/cm is the slab mass thickness and Es = 20 MeV is the scattering constant. Hence we have % x E 2 S s (ρr) xS 3X0 pβ

Since E Mμ, we can write pβ ≈ E and ﬁnally get for the beam broadening % 90 20 r 2 10 0.1cm. 3 × 13.3 3000

Exercise 3.2.15 The quantity z2 R/M (where z is the charge in e units, R the range and M the mass of the particle) is a universal function of βγ = p/M. As an example an α-particle having a kinetic energy Tα has the same range of a proton with kinetic energy Tp = Tα/4 2 (same βγ), because zα/Mα = 1/Mp. Exercise 3.2.16 Electrons having E = 1 GeV loose energy by bremstrahlung as −dE = E . dx brem X0

Therefore the mean electron energy after crossing a plate of thickness x is x E(x) = E0 exp − . (3.7) X0 Appendix: Solutions of Exercises and Problems 119

X0 is the aluminium radiation length whose inverse is

1 Z 2 ≈ D ln(183 Z −1/3) 3.8 × 10−2 cm2/g. X0 A

The radiated energy corresponds to the value of E(x) in Eq. (3.7)atx = ρd 13.5g/cm2, and then is x Eγ =E = E0 1 − exp − 0.40 × E0 = 400 MeV. X0

Exercise 3.2.17 (a) In vacuum muons travel along a circular orbit whose radius is

p[GeV/c] R[m]= 16.7m 0.3 B[T]

(b) Muons have an initial energy = 2 + 2 = E0 p0 m 511 MeV and a βγ equal to p/m 4.7. Hence we can assume for the energy loss rate in the gas a value close to that of a minimum ionizing particle dE MeV 2 . / 2 dx ion g cm

The energy lost after a complete round is approximately8 dE E ρ × 2π R 42 MeV dx ion and the ﬁnal energy is E1 = E0 − E 469 MeV. The corresponding muon = 2 − 2 momentum is p1 E1 m 457 MeV/c and so the radius of curvature after one round is R1 15.2m. Exercise 3.2.18 The scattered photon energy as a function of the scattering angle θ is

Eγ Eγ = 1 + (1 − cos θ)

8For an estimate it is not necessary to consider the actual trajectory of the muon which slightly differs from a circle. 120 Appendix: Solutions of Exercises and Problems where = Eγ /m 1, for 0.5 MeV photons. If E, T and m are the energy, kinetic energy and mass of the scattered electron, from the energy conservation we have

Eγ + m = Eγ + E and then (1 − cos θ) T = E − m = Eγ − Eγ = Eγ 1 + (1 − cos θ)

The maximum energy is obtained when the photon is scattered backward (θ = π) and hence we obtain for the Compton edge energy

2 2 T = Eγ Eγ 0.33 MeV max 1 + 2 3

Exercise 3.2.19 The energy of the Compton scattered photon as a function of θ is

Eγ Eγ = 1 + (1 − cos θ) where = Eγ /m. The electron kinetic energy is then

(1 − cos θ) T = E − m = Eγ − Eγ = Eγ 1 + (1 − cos θ)

This energy is maximum for θ = π and this value corresponds to the co-called ‘Compton edge’ 2 2Eγ Tmax = Eγ = Eγ . 1 + 2 m + 2Eγ

Solving the equation in Eγ we have √ Tmax + Tmax(Tmax + 2m) Eγ = . (3.8) 2

The three Tmax values shown in the ﬁgure are about 0.22, 0.62 and 0.80 MeV.Knowing that in γ -transitions, neglecting the nucleus recoil, Qγ = Eγ ,from(3.8) we get

Qγ (1) 0.37 MeV, Qγ (2) 0.81 MeV, Qγ (3) 1.0MeV

Exercise 3.2.20 (a) The muon velocity is Appendix: Solutions of Exercises and Problems 121

p p 10 β = = = √ 0.999944 E p2 + m2 102 + 0.1062

Thus Cherenkov effect is done because we have 1 1 β 0.999944 > = 0.99971 n 1.00029 (b) The Cherenkov opening angle is given by

1 1 cos θ = 0.99977 C nβ 1.00029 × 0.999944 corresponding to an angle of 1.2◦. (c) The number of Cherenkov photons per unit length in the visible bandwidth is α N /L ≈ z2 ω sin2 θ ≈ z2 750 sin2 θ cm−1, ph c C C where ω corresponds to the visible and near UV bandwidth, where Cherenkov radiation is possible (ω ≈ 2 eV). 10 GeV muons produced at 10 km reach the sea level since their mean decay length is (τμ 2.2 µs):

pμ 10 5 −6 lμ = βγcτμ = cτμ 310 2.210 km 62 km, mμ 0.106 and then emit Cherenkov photons along their whole pathlengths. For muons hitting normally the Earth surface (θZ = 0) we have

2 2 −1 6 5 Nph ≈ 750(1 − cos θC ) × L 750(1 − 0.99977 ) cm × 10 cm 3.410 .

For angles θZ > 0, the number of photons scales as sec θZ . Exercise 3.2.21 From the 4-momentum conservation in the Compton scattering we have

Eγ + m = Eγ + E k = k + p

where (Eγ , k), (Eγ , k ) and (E, p) are the 4-momenta of the incident photon, scattered photon and scattered electron respectively. We need to calculate the relationship between the initial photon and the scattered electron as a function of the electron angle φ. To get this we write

2 2 2 2 2 2 k = (k − p) = k + p − 2kpcos φ = Eγ + p − 2Eγ p cos φ

2 2 2 2 k = Eγ = (Eγ + m − E) = (Eγ − T ) 122 Appendix: Solutions of Exercises and Problems where T is the electron kinetic energy. Hence we have

p2 − T 2 Eγ = 2(p cos φ − T )

(a) An electron having an angle φ within the ﬁbre acceptance releases its whole kinetic energy, because it has enough pathlength to come at rest. Thus the measured energy release corresponds to the kinetic energy of the electron at φ = 30◦ and we have for the source energy

2.462 − 22 Eγ 7.9MeV, 2 · (2.46 · cos 30◦ − 2) where we have used p = (T + m)2 − m2 2.46 MeV/c. (b) The Klein–Nishina cross section is & ' 2 2 dσ r Eγ Eγ Eγ = 0 + − 2 θ sin (3.9) d 2 Eγ Eγ Eγ where θ is the photon scattering angle. To get this angle we equate the photon and electron transverse momenta

φ = θ =⇒ θ = p φ p sin Eγ sin sin sin Eγ

◦ The photon energy Eγ , corresponding to the electron emitted at 30 , can be derived from Eγ = Eγ − T . Hence we have Eγ /Eγ 0.75 and sin θ 0.21. Substituting these values in (3.9) we get

dσ 2.8fm2 × 0.752 × (0.75 + 1/0.75 − 0.212) 4.510−26 cm2/sr. d 2 The cross section for all the accepted electrons is this differential cross section multiplied by the acceptance solid angle 15◦ = 2π d cos θ = 2π(1 − cos 15◦) 6.28 × (1 − 0.966) 0.21 sr 0

Hence we have dσ σ 4.510−26 × 0.21 9.510−27cm2. acc d Appendix: Solutions of Exercises and Problems 123

Z μ = N ρσ 61023 × 0.5 × 1 × 9.510−27 0.0029 cm−1 acc A acc where we have taken into account that Z electrons per each atom contribute to the scattering. The number of detected electrons per incident photon is then . Ne d μ 0 2 × . . . ◦ acc 0 0029 0 001 Nγ sin 30 0.5

Exercise 3.2.22

By deﬁnition E(x) = E0 exp(−x/ X0), where X0 is the radiation length. The mean energy loss is then E = E0 − E(X0) = E0(1 − 1/e) = 0.63 GeV.

3.3 Detection Techniques and Experimental Methods

Exercise 3.3.1 (1) The inverse of β is given by

1 = 1 . β − 1 1 γ 2

Since E m,wehaveγ 1, so that can use the relationship, valid for x → 0

1 1 √ 1 + x2. 1 − x2 2

The difference between the time-of-ﬂights of two particles having velocities β1 and β2 is L L L 1 1 L 1 1 T = − 1 + − 1 − = − = β β γ 2 γ 2 γ 2 γ 2 1c 2c c 2 1 2 2 2c 1 2 L m m L m2 m2 L m2 − m2 = 1 − 2 1 − 2 = 1 2 , 2 2 2 2 2 2c E1 E2 2c p1 p2 2c p having set p1 = p2 = p in the last step. (2) The difference in time-of-ﬂight between pions and kaons is

L m2 3m 0.4932 − 0.1392 3m× 0.224 T = = 1.12 ns 2c p2 2 c 1 2 × 3 × 108m/s 124 Appendix: Solutions of Exercises and Problems

Using the time resolution requirement, T = 4σt , we obtain for the time resolution needed for each counter 1.12 ns σ √ 0.2ns, 4 2

σ 2 = σ 2( − ) = σ 2( ) + σ 2( ) = σ 2 where we have used the relationship t T1 T2 T1 T2 2 . (3) When S1 and S2 are segmented and a third scintillator S3 is inserted in the middle, the system can be used as a spectrometer. √ (a) The space resolution of each scintillator is σy=5cm/ 12 1.44 cm (y is the direction orthogonal to the beam in the ﬁgure). The lateral spread due to the multiple scattering (see Exercise 3.2.14), in the same direction, is: % % x Es x 1cm14MeV 1 0.013 σy √ √ √ mm, 3 2 pβ X0 3 1000 MeV β 40 β which turns out to be negligible for both particles (βπ 0.99, βK 0.90) with respect to the resolution. (b) The sagitta is

BL2 0.3 × 1T× 9m2 s = 0.3 = 0.337 m = 33.7cm. 8 p 8 × 1GeV/c

The sagitta is measured as s y3 − (y1 + y2)/2 and the uncertainty on each yi is σy. Hence we have for the sagitta uncertainty % 3 σ σ 1.76 cm. s 2 y

The relative error on the momentum measurement is ﬁnally

p s 1.76 cm = 5% p s 33.7cm

Exercise 3.3.2

The radius R of the orbit at t = to is

p[GeV/c] 0.3 R[m]= = m 2m. 0.3 B[T] 0.3 × 0.5

A 300 MeV/c muon (βγ 3) is at the minimum of the ionization loss rate. The medium is not speciﬁed but it may be a gas, considering its density. Let assume to be air for which the minimum ionization loss is ≈1.8MeVg−1 cm2. For the energy loss in iron we use instead ≈1.5MeVg−1 cm2. Under this assumption we have Appendix: Solutions of Exercises and Problems 125

MeV MeV E 1.5 × ρ × 2d + 1.8 × ρ × 2π R gcm−2 Fe Fe gcm−2 air

= 1.5 × 7.87 × 2 × 0.2 + 1.8 × 10−3 × 2π × 200 = 6.98 MeV.

The initial muon energy is

E = p2 + m2 3002 + 1062 MeV 318 MeV;

After one turn the energy becomes

E = E − E 318 − 6.98 MeV = 311 MeV. and the momentum is

p = E 2 − m2 3112 − 1062 = 292 MeV/c.

(a) The magnetic ﬁeld needed to keep the muon in an orbit of radius R after one turn is p [GeV/c] B = 0.486 T 0.3 R[m] and hence we have B = B − B 0.486 − 0.5 −0.014 T. (b) The muon mean decay pathlength is λ = βγcτ,where βγ = p/m. Hence we have p 300 × 3108 × 2.210−6 λ = cτ 1868 m. m 106 The mean number of turns is then λ 1868 n = 149. turns 2π R 4π Exercise 3.3.3 1. Muons come at rest in water loosing their kinetic energy by ionization. The energy lost by Cherenkov effect is negligible (order of per mill). Muons with 1 GeV/c momentum (m = 106 MeV/c2) are close to the ionization minimum and we can use an energy loss rate of about 2 MeV/(g cm−2)(= 2 MeV/cm in water). A simple estimate of the total pathlength can be done under the assumption that the changes in energy loss rate along the muon path can be neglected T dE T dE 900 MeV R(E) 4.5m, 0 (−dE/dl)ion 0 2MeV/cm 2MeV/cm where T = p2 + m2 − m 900 MeV is the initial kinetic energy of the muons. 126 Appendix: Solutions of Exercises and Problems

A better estimate is made using the range-versus-energy plots reported in the PDG Review of Particle Physics [1]. Here only a few elements are shown: in particular for 1 GeV/c muons we get R/m ≈ 2000 g cm−2 GeV−1 for H (Z/A = 1) and R/m ≈ 4000 g cm−2 GeV−1 for C (Z/A = 0.5). In the mixtures, as water, one has to take into account that the primary dependence of the ionization energy loss( is on the ratio / − / / = / = (Z A. Therefore( dE dx of the mixture is proportional to Z A w j Z j A j n j Z j / n j A j , where w j (n j ) is the weight fraction (number of atoms) of the j-th element in the compound. In water we have Z/A =(2 × 1 + 8)/(2 × 1 + 16) 0.56. Therefore the range in water is dominated by the energy loss in oxygen. If we take the range in carbon as a reference we obtain R 4000 cm/GeV × 0.106 GeV 4.2 m. We notice that the value obtained assuming a constant energy loss rate overestimates the actual range, but is adequate for a rough estimate. 2. The condition to emit Cherenkov photons is β ≥ βmin = 1/n 0.75. Hence we have for a particle mass m

pmin = βmin γmin m =⇒ pmin 1.134 m, and then for a muon = 2 + 2 − . . Tmin pmin m m 0 51 m 54 MeV

The length of the path where the muon emits Cherenkov radiation is (for a constant energy loss rate) T dE T dE 900 − 54 MeV L = . . C (− / ) / / 4 23 m Tmin dE dl ion Tmin 2MeV cm 2MeV cm

Comparing this value with the one obtained under the same approximation we obtain a fraction 4.23/4.5 94%. 3. The initial opening angle of the Cherenkov cone is obtained from

1 p2 + m2 1.006 cos θ = = 0.756. C β n pn 1.33

The muon energy loss up to the exit from the detector is small enough (≈100 MeV) so that the Cherenkov angle is almost constant. Hence the region illuminated on the base is determined by the Cherenkov cone at the initial point and the radius of the circle is

2 D 1 − cos θC R = D tan θC = 0.865 D 43 cm. cos θC Appendix: Solutions of Exercises and Problems 127

Exercise 3.3.4 Considering the distance and the size of the detector, electrons and positrons are detected for angles between

θmin = arctan(6/200) 30 mrad and θmax = arctan(10/200) 50 mrad. √ The energy of each beam is Ee = s/2 = 45 GeV. Integrating the given expression of the Bhabha cross section between θmin and θmax we have

θ 8πα2 max dθ 8πα2 1 1 σ = (c)2 = (c)2 − = 2 θ 3 2 · θ 2 · θ 2 Ee θmin Ee 2 min 2 max − 1 1 − − 2.57 · 10 8 fm2 − 0.91 · 10 5 fm2 9.1 · 10 32 cm2 2 · (0.030)2 2 · (0.050)2

For a rate of 1 ev/s we obtain a luminosity

n 1s−1 L = 1.1 · 1031 cm−2s−1 σ 9.1 · 10−32cm2 Exercise 3.3.5 If τ is the mean lifetime, the number of particles surviving after a time t is

−t/τ N(t) = N0e .

In our case we require to have at least one decay in t = 1yr t N − = N e−t/τ N − 0 1 0 0 1 τ

= τ = hence N0 t . Since the number of nucleons in the detector mass M is N0 NA M 23 −1 (where NA 6.02 · 10 mole is the Avogadro number) we obtain for the required mass τ 1032 M = 1.7 · 108 g = 170 ton 23 −1 tNA 6.02 · 10 g

Exercise 3.3.6 The pion momentum in GeV/c is given by the relationship

p = 0.3 BR, 128 Appendix: Solutions of Exercises and Problems where R is the curvature radius in metres and B the magnetic field in Tesla. The deflection angle in the magnetic field can be written to a good approximation as . θ L = 0 3 BL. R p

We notice that the approximation of using L equal to the length of the magnet (instead of the length of the trajectory) is justiﬁed by the fact that the deﬂection angles are small for all the momenta (130 mrad at 0.5 GeV/c down to 44 at 1.5 GeV/c). The distance L and the width w of the slit allow to select the pion momentum and its uncertainty. The angle subtended for a momentum p0 ± p is . θ w = 0 3 BL . 2 p d p0

Hence to select 1 GeV/c ± 5% charged pions we need a distance

p w 1 × 0.01 d = 0 1.5m . × . × . × . 0.3 BL p 0 3 0 2 1 1 0 1 p0

Exercise 3.3.7 The relationship for a particle of charge e among the curvature radius R in metres, the uniform magnetic ﬁeld B in Tesla and the momentum p in GeV/c is

p = 0.3 BR hence the muon momentum is p 2.1 GeV/c. For the revolution period we have

2π R T = , βc 2 2 where the velocity is β = p/E = p/ p + mμ ≈ 1. The revolution period is then T 2.93 × 10−7 s. The mean muon lifetime in the Lab system is p τLS = γτ τ 20 τ. mμ

The number of muons surviving after one period is

−T/τLS N(T ) = N0e , Appendix: Solutions of Exercises and Problems 129 where N0 is the initial muon number. The fraction of muons decayed after one period is then

− /τ N (1 − e T LS ) T T f = 0 1 − 1 − = 6.7 × 10−3. N0 τLS τLS

Exercise 3.3.8

1. The total energy of the particles involved in the proton decay is E0 = m p.In the considered decay channel the energies of the two particles are almost equal so that e ≈ π 0 ≈ E0/2 (the correct calculation gives 0.46 GeV and 0.48 GeV for the energies of the positron and pion respectively). Positrons and photons (from π 0 decay) are produced back-to-back and, having energies above the water critical energy and then produce e.m. cascades. Under the approximation of equal energies, the maximum of the longitudinal development is (in units of X0)

ln[E /(2E )] T = 0 c 2.55. max ln 2 Hence most of the Cherenkov emitting particles are contained in a segment of length = × X0 × . L 2 ρ Tmax 1 8 m. This length determines the size of the detector (each side L). 2. To estimate the number of emitted Cherenkov photons, we need to evaluate the total track length for the charged particles (e+, e−) contained each cascade. This total length (called track length integral) is given by 2 Tmax 2 2 E t Tmax 0 Ttot = 2 dt = (2 − 1) = − 1 4.7, 3 0 3ln2 3ln2 2Ec where the factor 2/3 is the average fraction of charged particles in the cascade. Hence we have for the total number of Cherenkov photons

X · T N = 0 tot × I × −1 . × 5 phot 2 ρ 0 338 cm 400 cm 1 4 10

Exercise 3.3.9 When a proton interacts with a residual air molecule it is thrown away from the trajectory where the accumulated protons are kept by the magnetic ﬁeld. Then at each scattering a proton is lost. The absorption coefﬁcient is given by

N μ = σnn= ρ A , A where σ is the total cross section, n is the number of scatterers per unit volume and −11 NA is the Avogadro number. In the proton ring (10 atm) we have 130 Appendix: Solutions of Exercises and Problems

61023 μ = 300 × 10−27 × 1.25 10−14 × 1.6 × 10−16 cm−1. 14 The inverse of this value corresponds is the mean pathlength. 300 GeV protons are ultra-relativistic and their velocity is ≈c. Hence the mean beam lifetime is

1 τ = 2.08 × 105 s 58 h cμ

Exercise 3.3.10 The interactions occurs against the nuclei along the beam. These are

N 6.02 × 1023 N = A ρ dS= × 11.3 × 0.2 × π × 1 2 × 1022 sc A 207 The fraction of scattered particle is given by

N 2 × 1022 f = s σ = × 310−26 1.910−4 s S 3.14 Exercise 3.3.11 The neutrino interaction rate is given by

−44 2 6 −2 −1 −38 −1 wint = σφ 7 × 10 cm × 10 cm s 7 × 10 s .

The number of scattering centres (electrons) per unit volume is

Z Z n = ρ N V = N M = 0.5 × 6.02 · 1023 × 5 · 1010 1.5 × 1034. sc A A A A

Hence the number of interactions per year is (T = 1yr 3.15 × 107 s)

4 Nyr = wint × nsc × T = 3.3 × 10

Exercise 3.3.12 (a) The maximum shower development is reached at a depth

log (E /E ) log (500 GeV/80 MeV) T = 10 0 c = 10 12.6 log10 2 log10 2 where T is expressed in radiation length units. Therefore the actual depth in g/cm2 is

2 Xmax = T × X0 470 g/cm corresponding to an optimal altitude (for vertical showers) Appendix: Solutions of Exercises and Problems 131 Xmax h =−h0 ln 5300 m. Xv(0)

There are sites suitable for such observations, e.g. in the Andes or in Tibet. (b) Electrons at the shower maximum have an energy equal to the critical energy water atm = Ec Ec 80 MeV. At this energy the Cherenkov condition is fulﬁlled

n 1.33 nβ = = > 1. 1 + (m/p)2 1 + (0.511/80)2

Hence Cherenkov photons can be used to detect shower events. (c) Shower particles (photons and electrons) at the critical energy have equal probability to loose energy by ionization and bremsstrahlung. Therefore they are not energetic enough to produce e.m. cascades. The component of the shower which are already electrons mostly loose energy by ionization. Instead those which are photons have still enough energy for pair production (threshold energy 1 MeV) and can generate electrons of both signs with lower energies. To make an estimate of the path done by electron loosing energy in water we can calculate the residual range of electrons at the critical energy

water Ec dE Ewater x = c = 40 g/cm2 −dE 2MeV/(gcm−2) 0 dx ion and then l 40 cm. Hence electrons loose their whole energies in the water tanks, apart those which exit the tank and loose only a part of their energy.

Exercise 3.3.13 1. The reaction threshold for the proton kinetic energy is

( + )2 − ( )2 2 2m p m J 2m p m J Tth = = 2m J + 11.3GeV 2m p 2m p

2. Denoting with M the total CMS energy for protons of 28 GeV energy against target protons (at rest), we have

M 2m p E p 7.3GeV.

The ﬁnal state in the reaction (3.2) is a three body system. Then the maximum and minimum energy of the J/ψ particle in the CMS are given by

: ∗ = . min E J m J 3 1GeV M2 + m2 − (2m )2 max : E∗ = J p 4.07 GeV. J 2M 132 Appendix: Solutions of Exercises and Problems

To obtain the maximum and minimum values in the Lab system we make a Lorentz transformation with the following β and γ values

p E + m β = p 0.967,γ= p p 3.96 E p + m p M

To calculate the minimum and maximum J/ψ energies in the Lab frame we consider the following cases = γ( ∗ +β · ) . min: E J E J [min] 0 12 3GeV = γ( ∗ −β · ∗ . max/min: E J E J [max] pJ [max] ) 6 0GeV = γ( ∗ +β · ∗ ) . max/max: E J E J [max]] pJ [max] 26 2GeV It follows that the minimum and maximum energies are 6 and 26.2 GeV respectively. + − 3. The minimum opening angle θmin of the e e pair is obtained from 2 − 2 θ m J 2me m sin min = J 2 E J E J

Therefore the minimum angle is obtained for the maximum J/ψ energy, 26.2 GeV, ◦ and turns out to be θmin 13.6 . 4. Electrons are ultra-relativistic (p E): hence the e+e− invariant mass is θ M2 4p+ p− sin2 ee 2 where p+(p−) is the e+(e−) momentum and θ is the opening angle of the observed pair. Using for Mee the J/ψ mass we obtain m2 + = J . / . p − 2 θ 12 4GeV c 4p sin 2

Exercise 3.3.14

(a) The collider system is the CMS, hence Eτ = E0/2 = 14.5 GeV. (b) Using the Sargent rue we have for the transition rates ( = 1/τ) (τ+ → + + ν +¯ν ) 5 e e τ = mτ . + + 5 (μ → e + νe +¯νμ) mμ

Taking into account the tau branching ratio into neutrinos we have

+ + + + BR(τ → e + νe +¯ντ ) (τ → e + νe +¯ντ ) = ττ

Hence the tau mean lifetime is (τ + → + + ν +¯ν ) 5 τ = BR e e τ = τ × (τ + → + + ν +¯ν ) × mμ τ + + μ BR e e τ (τ → e + νe +¯ντ ) mτ Appendix: Solutions of Exercises and Problems 133 106 5 2.210−6 × 0.18 × 3.0 × 10−13 s. 1777

2 − 2 pτ Eτ mτ 10 −13 L =βγcττ = cττ = cττ 8.1 × 310 × 310 0.073 cm. mτ mτ

For a cylindrical detector the minimum distance to observe a decay is given by the internal radius r (the distance increases with the angle). Hence the maximum detection probability is

∞ 1 l r − − −30 fmax(l > r) = e L dl = e L 210 L r and is then negligible.

Exercise 3.3.15 Photon and electron beams of 5 GeV produce electromagnetic showers. For their developments the relevant parameter is the number of radiation lengths. Each scintillator layer has 1/42 0.02 radiation lengths, whereas the lead slabs have about 2 radiation lengths each. Therefore the scintillator layers have a negligible contribution. Upstream of the fourth scintillator there are 3 lead slabs, hence the total number of radiation lengths is 3 × 1cm T = 5.4. 0.56 cm

Using the Heitler toy model the number of shower particles (e+, e− and γ )is2T .The scintillator detects charged particles via the ionization process whereas photons have a very low probability to convert to electrons because of the low Z of the material. In an e.m. showers charged particle are approximately 2/3 of the total content of particles, they have an energy loss rate corresponding to minimum ionizing particles ( 2MeVg−1 cm2) and then their total energy release is 2 dE E = × 2T × − × ρd 0.67 × 42.2 × 2 × 1.03 58 MeV, 3 dx ion where d is the scintillator thickness. This energy release is the same for incident electron and photons. Instead muons loose energy only by ionization. The energy lost before the fourth scintillator is 2 MeV g−1 cm2 × 3cm× 11 g/cm3 66 MeV, hence their energy is almost unaffected. We can assume for them the same energy loss rate of 2MeV g−1 cm2 and then the energy release in the fourth scintillator is

E = 2 × 1.03 2MeV. 134 Appendix: Solutions of Exercises and Problems

Finally to discriminate electrons from photons we can use the signal in the ﬁrst scintillator which can be detected only for electrons but is absent for photons. Exercise 3.3.16 (a) The p-Cu interaction length is

1 1 A A 3 63.5 3 λpCu = = = 18.5cm int 23 −27 NA ρσpCu NA ρσpp 610 8.96 40 10

(b) The initial state has baryon number B =+2, the two D-particles are mesons and have B = 0. Hence X must have B =+2. The simplest case is

p + p → D+ + D− + p + p

The flavor flux diagram is shown in Fig. 3.2 (left). (c) The quark flavor content of D+ and D− is D+ = cd¯ and D− =¯cd respectively. + + + + + D decays into neutrinos via c → W + s followed by W → l + νl . Hence D is associated to neutrinos. Similarly D− decays into W − and then anti-neutrinos are produced. Examples of Feynman diagrams with ν¯e and νμ final states are shown in Fig. 3.2 (right). (d) The interaction length of D-particles is

σpp 40 λDCu = × λpCu = × 18.5cm 25 cm σDp 30

The decay length is instead

± p ± λdec = βγ cτ(D ) = × cτ(D ) m D±

Therefore λdec λDCu is obtained for

λDCu 25 p m ± × 1.87 × GeV/c 1500 GeV/c D cτ(D±) 31010 × 1.04 10−12 which is always fulﬁlled for 400 GeV incident protons. (e) Taking into account the considerations at point (c) and the fact that BR(D± → ± νμ/ν¯μ) = BR(D → νe/ν¯e) we expect for the muon to electron neutrino ratio

νμ +¯νμ = 1. νe +¯νe

Exercise 3.3.17 The absorption coefﬁcient for pair production, which is the dominant process at high energies, is given by Appendix: Solutions of Exercises and Problems 135

Fig. 3.2 Flavor ﬂux diagram for p + p → D+ + D− + p + p (left). Feynman diagrams for two D decays (right)

7 μ = X −1 1.38 cm−1 9 0

A photon hitting the lead plate has a probability exp(−μd) to escape from the lead plate. Instead in case of pair production electrons emerge from the plate or induce e.m. showers, depending on the ﬁrst interaction point. If conversion occurs one or more electrons will reach the downstream detector. The conversion probability is then Pc = 1 − exp(−μd)

Hence the π 0 detection efﬁciency is

= 2 =[ − (−μ )]2 =[ − (− . )]2 Pc 1 exp d 1 exp 1 38 56%

Exercise 3.3.18 (a) The process that makes electron antineutrino detectable is the same used in the celebrated experiment by Reines and Cowan [2]

+ ν¯e + p → n + e

The process, called also ‘inverse beta decay’, is a charged current weak interaction (i.e. with W virtual boson). The detected particles are the positron through its annihilation in two photons of 0.5 MeV and the delayed photons emitted by the capture of the neutron. The process has the cross section given in the text. Instead muon antineutrinos, originating from the oscillation phenomenon, are difﬁcult to + detect. In fact the charged current process ν¯μ + p → n + μ is forbidden by kine- 0 matics (Ethr 100 MeV) and the neutral current process (i.e. with Z virtual boson) ν¯μ + p/n →¯νμ + p/n can be only detected from the nucleon recoil with very low 136 Appendix: Solutions of Exercises and Problems efﬁciency. Hence the oscillation phenomenon can be observed counting the number of disappeared electron antineutrinos. (b) If neutrinos do not oscillate the interaction rate is

Iν r = × σν × N × ρlS 4π L2 n where Nn is the number of target nucleons per gram, ρ is the medium density, l and S are the detector length and section respectively. The product ρlS is the detector mass and we have Iν r = × σν × N × M 4π L2 A

1018 s−1 61023 × 210−43 cm2 × × 106 g 2.410−5 s−1 12.56 200002 cm2 g

Denoting with the detection efﬁciency, the number of expected interactions per year is N = r × × T 2.410−5 × 0.70 × 3.15 107 529.

(c) For a detector at 200m from the reactor core and 2 MeV electron antineutrinos the probability to become muon antineutrinos is [ ] 2 −3 L m 2 −3 200 P(ν¯ →¯νμ) 0.20 sin 10 0.20 × sin 10 0.002 e E[MeV] 2

Hence the number of detectable electron antineutrinos is

Ne N ×[1 − P(ν¯e →¯νμ)]528 and the mean number of disappeared ν¯e is 1.1. (d) The probability to have a null result is given by the poissonian probability to observe no event out of an expectation of 1.1

1.10 P(0|2) = e−1.1 = e−1.1 33% 0! It is worth to notice that this probability is not realistic, because it is based on the assumption that the knowledge of the number of neutrino interactions is perfectly known. In real experiments the uncertainty on the neutrino ﬂux and detection efﬁ- ciency makes it impossible to observe a disappearance ratio (1/529) so small.

Exercise 3.3.19 (a) To get the mass of the particle we consider the region 2, after the slowing down, where two measurements are available. Appendix: Solutions of Exercises and Problems 137

. × 8 β From the time-of-ﬂight we have β = v2 2 8 10 0.93 and β γ = √ 2 2 c 3×108 2 2 −β2 1 2 2.60. From the curvature we have p2 = 0.3 BR2 = 0.3 × 1 × 1.21 0.363 GeV/c. The rest mass of the particle is

p 0.363 m = 2 0.140 GeV/c2. β2γ2 2.60

It is a charged pion whose momentum before the slowing down is

2 . 2 l1 0 80 p1 = 0.3 B 0.3 × 1 × 0.80 GeV/c, 8s1 8 × 0.03 and the kinetic energy is = 2 + 2 − . . T1 p1 m m 0 670 GeV

(b) The energy lost in the medium is = − = − 2 + 2 − . − . . . E T1 T2 T1 p2 m m 0 670 0 250 0 420 GeV

L / 14 τ = 1 2 2.6 × 10−8 s. 8 cβ2γ2 ln 2 3 × 10 × 2.6 × 0.69

Exercise 3.3.20 (a) Neglecting energy losses, the momentum of the electron (positron) is

p B R = 0.3 (3.10) GeV/c Tesla m

0.3 × 0.8 × 0.40 0.096.

The photon energy is the sum of the two momenta Eγ = 2 p 192 MeV. In this calculation the opening angle of the pair has not been considered: in fact it is negligible, −3 θ ≈ me/Eγ 2.7 × 10 . (b) To make a rough estimate of the energy loss along the electron (positron) track, we assume that the track length is the same as in the previous case (though the track actually changes). This energy loss is due to ionization, because the bremsstrahlung is negligible for E < Ecrit (≈300 MeV in LH2) 138 Appendix: Solutions of Exercises and Problems

Liquid H 2

e+ e-

R1 R2 γ

Fig. 3.3 Solid line: no energy losses; dashed line: with energy losses dE E − ρπR dx ion

9 Since the electron (positron) has p/me of few hundreds, we can assume that −1 2 (−dE/dx)ion ≈ (−dE/dx)min 4.1MeVg cm . Therefore we have

E 4.1 × 0.071 × 125.6 36.6MeV.

The electron (positron) momentum at the entrance of the chamber can be estimated using the following arguments (see Fig.3.3):

• the track is not a semi-circle: its radius at entrance R1 (at exit R2) is larger (smaller) than the circle radius in the case of no energy losses R; • the sum of these two radii can be approximated to 2R (the measured diameter).

Since R ∝ p, denoting with pin and pout respectively the momentum at entrance and exit of the chamber, from 2R = R1 + R2 we obtain p + p 2p − p p = in out = in . 2 2 Therefore we have: p E 37 p p + p + 96 + 114.5MeV in 2 2 2

corr and the photon energy is Eγ = 2 pin 229 MeV. A more accurate calculation can be done as follows. dE dE dp − ρdl − ρ Rdα. dx ion dx ion

9This value can be obtained from either the table “Atomic and nuclear properties of materials” in the PDG Review of Particle Physics [1] or simply from a standard value (for light materials) of 2MeVg−1 cm2 multiplied by 2 because of the Z/A ratio for the Hydrogen. Appendix: Solutions of Exercises and Problems 139

In this expression we have assumed that the arc element is centered as in the case of no losses: it is not actually true, but it is a sensible approximation for an estimate. Substituting here Eq. (3.10) one gets dp dE 3 = − ρ dα =−kdα, p dx ion B with k = 4.1 × 0.071 × 3/0.8 0.121. Integrating we have

p(α) = pin exp(−kα).

Therefore p = pin − pout = pin [1 − exp(−kπ)]

p 37 p = 117 MeV in 1 − exp(−kπ) 1 − exp(−0.121π)

corr from which we get Eγ = 2 pin 234 MeV.

Exercise 3.3.21 (a) The dominating process at this energy is Compton scattering by which photons transfer part of their energies to electrons. Iterating this process the whole energy of the photons is deposited and the measurement is possible through the ionization energy loss of the electrons. The characteristic length which is relevant to determine the sizes of the detector is the Compton mean free path

A 1 λC = , Z NA σC where σC is the Compton cross section, NA is the Avogadro number and A, Z refer to the detector material. To make a rough estimate one can assume A/Z ≈ 2 and use the Thomson cross section, σT , for the Compton scattering

1 λ ≈ 2 × 5gcm−2. C 6 × 1023 × 6.6 × 10−25

A more accurate calculation would give a larger λC (by about a factor 2), being the Thomson cross section the low energy limit of Compton scattering. + (b) In the antineutrino scattering ν¯e + p → e + n, the outgoing particles have momenta of the same order of the momentum of the incident neutrino. Assuming that pn ≈ Eν , the neutron is non-relativistic and we get for the its kinetic energy

2 Eν Tn ≈ 2keV 2 mn 140 Appendix: Solutions of Exercises and Problems which means that the recoil energy is negligible with respect to the other energies.10 (c) Denoting by E+ the positron energy, from energy conservation in the process + ν¯e + p → e + n, neglecting the neutron recoil energy, we get

Eν = E+ + mn − m p.

Energy conservation applied to the positron annihilation gives

Evis = E+ + me. (3.11)

Then the asked relationship is

Eν = Evis + mn − m p − me Evis + 0.78 MeV

(d) Since Evis ≥ 2me because of Eq. (3.11), the detected neutrinos must have

Eν ≥ 2me + 0.78 MeV 1.78 MeV.

This corresponds to the energy threshold of the process.

References

1. Tanabashi, M., et al.: (Particle data group). Phys. Rev. D 98, 030001 (2018). http://pdg.lbl.gov/ 2. Reines, F., Cowan, Jr., C.L.: Free anti neutrino absorption cross section. I. Measurement of the free anti neutrino absorption cross section by protons. Phys. Rev. 113, 273 (1959)

10The accurate calculation can be done in a speciﬁc case, e.g. the maximum kinetic energy: it is obtained for a neutron emitted in the forward direction. In this case we have all momenta along the same direction. Note that the calculation requires to use accurate mass values for the three 2 particles: m p = 938.272, mn = 939.565, me = 0.511, all in MeV/c . The invariant mass squared 2 = 2 + . is M m p 2 m p Eν and then M 940 27 MeV. The neutron energy in the CMS is

2 2 ∗ M + m E = n 939.565 MeV, n 2 M ∗ and the corresponding momentum is p 0.48 MeV/c. The β of the CM is β = Eν /(Eν + m p) 0.021 and then the Lorentz factor is 1. The neutron momentum in the Lab is = γ( ∗ + β ∗) . + . × . . / , pn p En 0 48 0 0021 939 565 2 5MeV c which corresponds to a neutron kinetic energy 3.2 keV.