Part II Particle and Nuclear Physics Outline Solutions

David Ward Lent/Easter Terms 2014

Part II Particle and Nuclear Physics Outline solutions

1. Natural Units; practice in using natural units, and converting between natural and SI units

(a)

1 1 λ = =7.15 GeV− =1.41 fm m usinghc ¯ =0.197 GeV fm to change units. (b) We have α 1/137 and √s =91.2 GeV. ≈ 8 2 8 2 11 σ =2.68 10− GeV− =2.68 10− (0.197 fm) =1.04 10− barns = 0.0104 nb ⇒ · · × · (c) Using standard dimensional analysis to insert the missing factors ofh ¯ and c we ﬁnd:

h¯ 4πα2h¯2c2 λ = and σ = mc 3s (noting that α is dimensionless, so equal to 1/137 in any system of units).

2. Semi-Empirical Mass Formula

(a) Bookwork. 1 (b) Treat nucleus as uniformly charged sphere of radius R = R0A 3 , charge Ze. Find self- energy by bringing spheres of charge from inﬁnity. Charge of sphere of radius r is 3 2 4 3 Ze(r/R) , and charge of spherical shell between r and r + dr is Ze(4πr dr)/( 3 πR ). So, the self-energy is R 4πr2dr Ze(r/R)3 3Z2e2 Ze 4 3 = Z0 3 πR 4πǫ0r 20πǫ0R Hence, 3e2 aC = =0.72 MeV 20πǫ0R0 (c) The B/A vs A curve peaks around A 56, so for light nuclei fusion can be energetically favourable, while for very heavy nuclei∼ ﬁssion is possible.

1 (d) At ﬁxed A the most stable nucleus will be that with lowest mass (not necessarily the ∂M greatest binding energy). So we require =0 ∂Z !A

2aC Z 4aA(A 2Z) (mn mp)+4aA (mp mn)+ 1 − =0 Z = − 1 ⇒ − A 3 − A ⇒ 2aC A− 3 +8aA/A Inserting numerical values for A = 300 gives Z = 114 etc. (e) By analogy with (b), and neglecting the proton-neutron mass diﬀerence, we have the gravitational p.e. given by:

2 2 2 2 3GM 3Gm A 5 3Gm n 3 n 37 = 1 = aGA with aG = =5.8 10− MeV − 5R − 5R0A 3 − 5R0 · If neutrons only, N = A and Z = 0, so the binding energy reduces to

2 5 B = a N a N 3 a N + a N 3 V − S − A G which must be > 0 if the ”nucleus” is bound. The aS term is negligible, so we ﬁnd 2 N 3 > (a a )/a =5 1055. So M > 8 1028 kg which is 0.07 solar masses. A − V G · · 3. The Fermi Gas model Density of states: 3 2 3 1 4√2m R0 g(ǫ)= BAǫ 2 where B = 3πh¯3 Fermi energy is deﬁned by

2 ǫF 3 3 2 2 3N N = g(ǫ)dǫ = BAǫF ǫF = Z0 3 ⇒ 2BA 1 If N = Z = 2 A, neutrons and protons have the same Fermi energy

2 2 3A 3 3 3 ǫF = = = 33.2 MeV 4BA 4B

(n.b. non-relativistic) and the corresponding Fermi momentum = √2mǫF = 250 MeV/c. Total KE carried by the neutrons is

5 2 ǫF 2 5 2 3N 3 3 3 3 5 2 3 ǫg(ǫ)dǫ = BAǫF = BA = N Z0 5 5 2BA 5 2BA Adding in the protons similarly, total KE is

2 3 3 3 5 5 (N 3 + Z 3 ) 5 2BA Writing N = 1 A(1 + α) and Z = 1 A(1 α) where α =(N Z)/A, we can expand to second 2 2 − − order in α 5 2 5 5 A 3 10 (N Z) 3 3 (N + Z ) 2+ −2 ≈ 2 " 9 A #

2 So the asymmetry term in the energy is

2 5 3 3 3 A 3 10 (N Z)2 (N Z)2 −2 aA − 5 2BA 2 9 A ≡ A 1 from which we ﬁnd, after a little algebra, aA = 3 ǫF = 11 MeV. This is about half the ﬁtted value for aA. Pairing energy 1/g(ǫ ) in this model. ∼ F 1 2 3A g(ǫF )= BAǫF = 4ǫF

so pairing energy 4ǫF /3A 0.44 MeV, taking A = 100 and ǫF = 33 MeV. For comparison, ∼ ∼3 SEMF ﬁt gives 33.5 MeV/A 4 1 MeV. Again the Fermi gas model is smaller by about a factor of two, because it only takes∼ account of KE, and not of any PE contribution, i.e. the inter-nucleon attraction. 4. Practice in using cross-sections

2 (a) Number of scattering centres per unit area = ρ/MA where ρ = 0.1 kg m− , and MA 5 1 is the mass per nucleus. Rate of scattering = 10 (ρ/MA)σ = 688 s− (since the total 1 cross-section σ = 270.01 b); transmitted intensity = 99312 s− . 1 (b) Rate of ﬁssions = 688 (200/270.01) = 510 s− . × 1 (c) Total rate of elastic scattering = 688 (0.01/270.01) = 0.0255 s− . Area of sphere at 2 × 5 1 2 radius 10 m is 400π m , so ﬂux of neutrons = 0.0255/400π =210− s− m− . 5. Nuclear sizes and Form Factors Start from 2 3 iq r F (q )= d r e · ρ(r) Z For the purpose of integration, take q as the polar axis, hence:

2 2 iqr cos θ′ F (q ) = 2πd(cos θ′) r dr e ρ(r) Z ′ π eiqr cos θ 4π = 2πr2drρ(r) = ∞ r sin qrρ(r)dr Z " iqr #0 q Z0 Then, for qr 1, expand the sin keeping the ﬁrst two terms: ≪ 4π (qr)3 1 F (q2)= ∞ r qr ρ(r)dr =1 q2R2 q Z0 " − 3! ···# − 6 ···

15 1 For a 200 MeV electron (ultrarelativistic) E = p = k in natural units, so k =1.02 10 m− . 1 14 1 2 2 2 ×15 q =2k sin θ =1.95 10 m− . So, given F (q ) =0.7 we obtain √R =5 10− m. 2 × | | × Oscillatory structure in F (q2) suggests non-smoothness in ρ(r), i.e. more like a top-hat than a Gaussian form (c.f. Fourier transforms of a Gaussian and a top-hat in 1D.) Best ﬁt is a top-hat with rounded corners (e.g. a Fermi function).

3 6. Mirror Nuclei 1 1 The pair of mirror nuclei have (N,Z)=( 2 (A 1), 2 (A 1)). They have different Coulomb energies; ± ∓ 2 2 aC A +1 A 1 aCA 3αA ∆EC = 1 − = 1 = A 3 " 2 − 2 # A 3 5R 2 1 using aC =3e /(20πǫ0R0)=3α/5R0 and R = R0A 3 . The only other terms which are differ in the SEMF between the mirror nuclei are Z(m +m )+(A Z)m , so M(A, Z +1) M(A, Z)= p e − n − 2me + Emax = ∆EC + mp + me mn ∆EC = Emax +1.8 MeV. Insert numbers to get the − ⇒ 1 answers on the problem sheet. Find R/A 3 is reasonably constant, but higher than the usual value of R0. 7. Scattering in Nuclear Physics Born approximation is based on first order perturbation theory, so it is suitable when the potential is weak. In contrast, the partial wave method is general, but involves the calculation of an infinite number of partial waves. However, if the projectile’s energy is small enough, and/or the potential is sufficiently short range, so that kR 1, then only the ℓ = 0 partial ≪ wave is significant in the neighbourhood of the potential, and the method is quite easy to apply. Of course, if the potential is also weak, then either approach is viable. The question doesn’t imply that V (r) is weak, but we are asked about the low energy limit, so the partial wave approach is the obvious choice. Wave functions are:

2m(V E) 2mV rb ψ = B sin(kr + δ0) where k = s h¯2 Matching boundary conditions at r = b we have:

A sinh qb = B sin(kb + δ0) ; qA cosh qb = kB cos(kb + δ0)

Dividing: tanh qb tan(kb + δ ) kb + δ = 0 0 as k, δ 0 q k ≈ k 0 → Hence: δ tanh qb dσ tanh qb 2 f(θ) 0 = b = b − ≡ k q − ⇒ dΩ " q − # 8. Breit-Wigner Formula in Nuclear Scattering We have Γn = 0.0104 eV and Γγ = 0.105 eV, so that Γ = Γn +Γγ = 0.1154 eV, assuming that there are no other decay modes. Hence, σn/σγ =Γn/Γγ, and given σγ = 75 nb, we obtain σn =7.4 kb. The c.m. energy of the neutron is approximately equal to the lab. energy of 2.2 eV, since the 2 11 target is much more massive. We thus infer λ = h /2mE = 1.92 10− m. We then know × everything in the cross-section formula apart fromqg (note that E = E0 at the resonance peak). We ﬁnd g=0.77 3 , which corresponds to J = 1 for the spin of the compound nucleus. ≈ 4 4 9. Nuclear Shell Model Sequence of energy levels is given in lecture notes. Each has degeneracy (2j + 1). Shell model predicts that the spin and parity is that of the unpaired nucleon. Hence:

Z N Unpaired nucleon predicted J P c.f. experiment? 3 1 + 2He 2 1 1s1/2 2 yes 9 3 − 4Be 4 5 1p3/2 2 yes 7 3 − 3Li 34 1p3/2 2 yes 12 + 6 C 66 None 0 yes 13 1 − 7 C67 1p1/2 2 yes 15 1 − 7 N78 1p1/2 2 yes 17 5 + 8 O89 1d5/2 2 yes 23 5 + 11Na 11 12 1d5/2 2 no 131 11 − 54 Xe 77 54 1h11/2 2 no 207 5 − 82 Pb 125 82 2f5/2 2 no

23 The last three disagree with experiment. 11Na has a signiﬁcant electric quadrupole moment, which tells us that the nucleus is not spherical, unlike the assumption in the simple shell model. 131 207 In the case of 54 Xe and 82 Pb, levels 2d3/2 and 3p1/2 respectively are predicted to lie very close to the naively predicted level, which would explain the observed J P values. It is often favourable to ﬁll a higher-j level at the expense of a lower-lying lower-j level, because the pairing energy is greater for high j.

10. Magnetic moments of nuclei Method – ﬁnd the state of the unpaired nucleon, and use the Schmidt limit formulae from notes: µ = gJ µN mJ for mJ = J where g g 1 g = g s − ℓ for the cases j = ℓ J ℓ ± 2ℓ +1 ± 2

and gs = 5.58 gℓ = 1 for an unpaired proton and gs = 3.83 gℓ = 0 for an unpaired neutron. Hence we have: −

3He unpaired n in 1s gives µ = 1.91µ 2 1/2 − N 9Be unpaired n in 1p gives µ = 1.91µ 4 3/2 − N 7 3Li unpaired p in 1p3/2 gives µ =3.79µN 13 6 C unpaired n in 1p1/2 gives µ = +0.64µN 15N unpaired p in 1p gives µ = 0.26µ 7 1/2 − N 17O unpaired n in 1d gives µ = 1.91µ 8 5/2 − N 9 Some correlation with experiment, but by no means perfect agreement. Least successful are 4Be 7 and 3Li which have a partially ﬁlled 1p3/2 — in this case the simple shell model usually predicts correctly the largest component of the wavefunction, but there are other ways of coupling the

5 angular momenta of the nucleons, which presumably contribute. The other four nuclides are all closed shell 1 nucleon, and the model is closer to reality. ± 3 Harder part: The d + p model of 2He is an example of an additional contribution to the wavefunction, which could still have the same J P as the shell model predicts. We can work out the magnetic moment for this case by noting that there are still only two contributions 1 to the total angular momentum, the spin of the proton (= 2 ) and the spin of the deuteron (=1). We can therefore adapt the regular formula by replacing ℓ by Sd = 1 and gℓ by gd = 1 2 (gp +gn)=0.875. This gives µ = 0.35µN , much worse than the shell model. The shell model, assuming like nucleons pair whenever− possible, may not give exactly the right wave-function, but it generally gives the largest term in the wavefunction correctly.

11. Spin-parity of Energy Levels

5 (a) Need to form an antisymmetric wave function for three j = 2 neutrons, so they must have mj values. Tabluate the possibilities and the corresponding values of total MJ :

mj values MJ 5 3 1 9 ( 2 , 2 , 2 ) 2 ( 5 , 3 , 1 ) 7 2 2 − 2 2 ( 5 , 3 , 3 ), ( 5 , 1 , 1 ) 5 2 2 − 2 2 2 − 2 2 ( 5 , 3 , 5 ), ( 5 , 1 , 3 ), ( 3 , 1 , 1 ) 3 2 2 − 2 2 2 − 2 2 2 − 2 2 ( 5 , 1 , 5 ), ( 5 , 1 , 3 ), ( 3 , 1 , 3 ) 1 2 2 − 2 2 − 2 − 2 2 2 − 2 2

9 7 plus similar negative values of MJ . There’s only one way to form MJ = 2 , 2 , but two ways 5 3 1 9 5 3 to make MJ = 2 and three ways to make MJ = 2 , 2 . Need J = 2 , 2 , 2 (only) to account for this.

(b) Phonons are bosons, so all combinations of mj values are possible. Tabluate these for two phonons:

mj values MJ (2,2) 4 (2,1) 3 (2,0), (1,1) 2 (2,-1), (1,0) 1 (2,-2), (1,-1), (0,0) 0

plus similar negative values of MJ . We see these form J = 4, 2, 0. Likewise for three phonons:

6 mj values MJ (2,2,2) 6 (2,2,1) 5 (2,2,0), (2,1,1) 4 (2,2,1), (2,1,0), (1,1,1) 3 (2,2,-2), (2,1,-1), (2,0,0), (1,1,0) 2 (2,1,-2), (2,0,-1), (1,1,-1), (1,0,0) 1 (2,0,-2), (2,-1,-1), (1,1,-2), (1,0,-1), (0,0,0) 0

plus similar negative values of MJ . The number of possible combinations changes at MJ values of 6, 4, 3, 2, 0 indicating that these are the permitted values of J. 12. Energy Levels; excited states of nuclei P + 18 Shell model predicts that the ground state of even-even nuclei should have J =0 . Only 9 F is not even-even, so it must be (d). 18 18 8 O and 10Ne are mirror nuclei, so one expects similar level schemes. They must be (b) and (e). 18 The difference between them is the Coulomb energy, which will be greater for 10Ne. To first order this has been subtracted out by measuring the energies w.r.t. the ground state. However, the Coulomb repulsion is slightly less for excited states than for the ground state, because their wavefunctions will be spatially larger, and we have therefore overcompensated for their 18 Coulomb energy by aligning the ground states. The upshot is that the excited states for 10Ne 18 18 will lie lower w.r.t. the ground state. Hence, 10Ne is (b) and 8 O is (e). This subtle effect is called the Thomas-Ehrman shift. 208 82 Pb is doubly magic, so we expect it to be very stable with a large first excitation energy — it must be (a). 166 By elimination, 68 Er must be (c). Its low-lying excitations clearly form a series of rotational states with energy J(J + 1). This is as expected for a heavy nuclide far from closed shells, which is quite likely∝ to be non-spherical. 13. Rate equations

197 198 (a) Denote the number of Au nuclei by N1 and the number of Au nuclei by N2. We then have: N˙ = R λN 1 − 1 N˙ 2 = λN1

6 1 where the decay rate is the reciprocal of the mean lifetime, λ = 2.89 10− s− , and 10 1 × R = 10 s− . Integrating, we obtain

R λt 15 N = 1 e− =2.69 10 for t = 6 days 1 λ − × Further integration gives

λt e− 1 15 N2 = R t + =2.50 10 " λ − λ# ×

7 197 The equilibrium number of Au nuclei is given by the limit of N1 for large t, which is given by N = R/λ =3.46 1015. 1 × (b) The rate equations are:

λCst N˙ = λ N N (t)= N (0)e− Cs − Cs Cs ⇒ Cs Cs λCst N˙ = λ N λ N = λ N (0)e− λ N Ba Cs Cs − Ba Ba Cs Cs − Ba Ba

This can be solved by various standard methods (e.g. integrating factor; complementary function + particular integral), noting the initial condition NBa(0) = 0 to yield

λCsNCs(0) λCst λBat N (t)= e− e− Ba λ λ − Ba − Cs h i The Ba activity (i.e. its rate of decay, λBaNBa(t)) is maximised when NBa(t) is maximised. i.e.

λCst λBat ln(λCs/λBa) λ e− = λ e− t = = 33.5 min Cs Ba ⇒ (λ λ ) Cs − Ba Activity of Ba relative to initial activity of Cs is

λBaNBa(t) λBa λCs λCst λBat = e− e− =0.087 λ N (0) λ λ λ − Cs Cs Cs Ba − Cs h i so the maximum activity is 87 µCi. 14. Alpha decay

(a) The likely decay scheme is this (not to scale):

200X 1−

α 4.650 α 4.687

γ 0.305 γ 0.266 196Y 0−

Note the energies of the α and γ almost add up to the same total, but not quite. This is because of the small recoil kinetic energy carried by the daughter nucleus, which is slightly diﬀerent in the two decay chains. P + The α-particle has J = 0 (even-even nucleus). The decay 1− 0− would require orbital angular momentum ℓ = 1 in order to conserve total angular momentum,→ but then 196 parity wouldn’t be conserved. The excited states of Y could be 1−, with ℓ = 0 in the α-decay, followed by M1 photon transition to the ground state, or 1+, with ℓ = 1 in the α-decay, and E1 transition to the ground state, or 2+, with ℓ = 1 in the α-decay, and E2 transition to the ground state.

8 (b) Geiger-Nuttall tells us that (for ﬁxed Z as in this case):

1 ln τ 1 = a + bQ− 2 2

1 For the Thorium isotopes given, a plot of ln τ 1 against Q− 2 shows good linear behaviour. 2 One can thus use the graph (or a ﬁt to the graph) to interpolate and determine τ 1 for 2 224 90 Th using the given value of Q. 15. Nuclear β-decay Main ingredients: Fermi transitions have S(eν) = 0 and Gamow-Teller have S(eν) = 1. Allowed have ℓ = 0 and ”nth-forbidden” have ℓ = n. Then apply usual QM rules for combination of angular momenta. Odd ℓ leans to a parity change, even ℓ does not. The lowest permitted value of ℓ will dominate.

1 + 1 + (i) 2 2 is consistent with either Fermi or G-T transitions — can couple either S(eν)=0 → 1 1 or S(eν)=1to 2 and get 2 . Superallowed, because overlap between nucleon wavefunctions must be maximal. (ii) 0+ 1+ is consistent with ℓ = 0 and S(eν) = 1 (but not S(eν) = 0). So pure G-T and superallowed→ (based on the ft value). (iii) 0+ 0+ is consistent with ℓ = 0 and S(eν) = 0 (but not S(eν) = 1). So pure Fermi and superallowed→ (based on the ft value). 3 + 3 + (iv) 2 2 is consistent with either Fermi or G-T transitions — can couple either S(eν)=0 → 3 3 or S(eν)=1to 2 and get 2 . Allowed (based on ft). + (v) 2− 0 has a parity change, so ℓ is odd. Consistent with ℓ = 1 and S(eν) = 1 (but not S(eν→) = 0). So pure G-T and ﬁrst forbidden. + (vi) 1− 0 has a parity change, so ℓ is odd. Consistent with ℓ = 1 and S(eν) = 0or S(eν→) = 1, so mixed Fermi and G-T, ﬁrst forbidden. 7 + 3 + (vii) 2 2 has no parity change, but ℓ = 0 does not work. Hence ℓ = 2 with either S(eν→)=0or S(eν) = 1, so mixed Fermi and G-T, second forbidden.

16. Nuclear β-decay Here, as well as considering selection rules, as in the preceding question, we also have to consider which decays are energetically possible. In terms of atomic masses, the relevant constraints are:

β+ : M(Z) M(Z + 1) > 0 − β− : M(Z) M(Z + 1) 2m > 0 − − e ElectronCapture : M(Z) M(Z + 1) > 0 −

Based on this, the Ce and Nd nuclides should be stable. The possible decays are:

+ La Ce by β− decay. 2− 0 implies first forbidden G-T (c.f. Qu. 15(v)). • → → + Pr Ce by EC. 2− 0 implies first forbidden G-T. • → → 9 + Pr Nd by β− decay. 2− 0 implies first forbidden G-T. • → → Pm Nd by EC or β+ decay. 1+ 0+ implies allowed G-T (c.f. Qu. 15(ii)). • → → Sm Pm by EC or β+ decay. 0+ 1+ implies allowed G-T. • → → The shortest lifetime will correspond to the allowed decay with the greatest energy release, i.e. Pm. The longest lifetime will correspond to the most forbidden decay with the smallest energy release, i.e. Pr. Agrees with experiment. 17. Gamma decay; selection rules P 3 + We are told the ground state is J = 2 . γ3 is an E2 transition to the g.s., so ∆J =2 and no P 1 + 3 + 5 + 7 + parity change. This is consistent with the 2.51 MeV state having J = 2 , 2 , 2 or 2 . But the first three of these could also decay to the g.s. by M1 transition, not seen. So the 2.51 MeV P 7 + state has J = 2 .

γ1 is M1, so ∆J = 1 and no parity change. This is consistent with the 2.10 MeV state having P 5 + 7 + 9 + 5 + 7 + J = 2 , 2 or 2 . But if it was 2 or 2 it could decay to the g.s. by E2 (or M1 in the latter P 9 + case), not seen. So the 2.10 MeV state has J = 2 .

γ2 is E1, so ∆J = 1 and a parity change. This is consistent with the 1.20 MeV state having P 5 − 7 − 9 − 5 − J = 2 , 2 or 2 . But if it was 2 it could decay to the g.s. by E1, which is not seen. So the P 9 − 7 − 1.20 MeV state has J = 2 or 2 (which can decay to the g.s. by the higher order processes E3 or M2/E3 respectively). 18. Nuclear Fission

(a) Suppose initial neutron (momentum p0, mass m) hits nucleus (mass Am where A = 12 in this case) head on. Final particles have momenta p1 and p2 in the same direction. We then have: p0 = p1 + p2 (conservation of momentum) p2 p2 p2 0 = 1 + 2 (conservation of energy) 2m 2m 2Am

Eliminate p2 and solve for p1/p0 we ﬁnd p 1 A 11 1 = − = p0 1+ A −13 Hence the maximal fractional energy loss for the neutron in one collision is 1 (11/13)2 = 48 − 169 . The smallest number of collisions, n, to reduce the energy to 0.025 eV is therefore given by 11 2n 0.025 = n = 55 13 2.5 106 ⇒ × This requires n consecutive head-on collisions, which is highly unlikely, so this is a con- siderable underestimate.

(b) Call the numbers of each nuclide present NU , NC and NB in obvious notation. The fraction of 235U by weight is 235N f = U 235NU + 12NC

10 where the contribution of B is negligible in f. We therefore have

NU 12f NB 12 6 6 = ; = 10− =1.2 10− N 235(1 f) N 10 × × C − C The probability that a neutron is absorbed by a 235U nucleus and generates ﬁssion is therefore N σ NU 580 P = U U;ﬁss. = NC × NU 6 NC σC + NU σU + NBσB 0.04 + 700+1.2 10− 3800 NC × × × NU 5 The multiplication factor is given by k = 2.5P < 1, which leads to < 5.94 10− NC 3 × which in turn implies f < 1.16 10− . × 1 19. Nuclear Fusion The two oxygen nuclei each have radius R 1.2 fm A 3 = 3.0 fm. The ∼ × Coulomb barrier to fusion is therefore the potential energy when the nuclei are 2R apart, i.e. 2 2 2 Z e Z α 1 = =0.08 fm− = 15 MeV. 4πǫ 2R 2R 0 × Setting this equal to k T gives T =1.7 108 K. B × 20. Relativistic Kinematics

(a) Conservation of energy: mX = Ea + Eb Conservation of momentum: E2 m2 = E2 m2 a − a b − b q q Eliminating Eb we have 2 2 2 2 2 2 2 2 2 mX + ma mb Eb = mX + Ea 2mX Ea = Ea ma + mb Ea = − − − ⇒ 2mX

and the result for Eb follows by interchanging a and b. If ma = mb both reduce to 1 Ea = Eb = 2 mX . (b)

m4 + m4 + m4 2m2 m2 2m2 m2 2m2m2 p = E2 m2 = X a b − X a − X b − a b = p a a − a q 2m b q X If m = m this reduces to p = p = 1 m2 4m2 a b a b 2 X − a (c) c.m. energy √s is given by q s =(E + E )2 (p + p )2 (920 + 27.5)2 (920 27.5)2 √s = 318.1 GeV e p − e p ≈ − − ⇒ This is suﬃcient energy to produce a Higgs boson. The case of ﬁxed-target collision can be obtained from the same formula, setting E = m and p = 0, giving s 2E m , from p p p ≈ e p which we obtain Ee = 53.9 TeV.