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Appendix Solutions of Exercises and Problems Appendix Solutions of Exercises and Problems 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 oppo- site 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 8 6 4 2 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 (kine- matical, 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 = .
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