8.701 Introduction to Nuclear and Particle Physics, Problem Set 5 Solutions

Massachusetts Institute of Technology Department of Physics Problem Set 5 handed out November 4th, 2020 Problem 1: The T2K Experiment[50 points] The T2K experiment uses an off-axis νµ beam from π + → µ+ νµ decays. Consider the case where the pion has velocity β along the z-direction in the laboratory frame and a neutrino with energy E ∗ is produced at an angle θ∗ with respect to the z 0 -axis in the π + rest frame. (a) Show that the neutrino energy in the pion rest frame is p∗ = (m2π − m2µ )/2mπ . • In the pion rest frame, the momenta of the particles are: p∗π = (mπ , ~0) q ∗ pµ = ( p∗2 + m2µ , p~∗ ) p∗ν = (p∗2 , −p~∗ ) 1 So, we have: p∗2 = (pµ∗ + pν∗ )2 π q 2 ∗ mπ = (p + p∗2 + m2µ )2 q m2π = 2p∗2 + 2p∗ p∗2 + m2µ + mµ2 q 2 2 ∗ ∗ mπ − mµ = 2p (p + p∗2 + m2µ ) m2π − mµ2 p 2(p∗ + p∗2 + m2µ ) p∗ = But, p∗π = pµ∗ + pν∗ ⇒ mπ = p∗ + p ∗2 p + m2µ . Therefore, p∗ = mπ2 − mµ2 . 2mπ (1) (b) Show that the energy E and angle of the production θ of the neutrino in the laboratory frame are E = γE ∗ (1+β cos θ∗ ) and E cos θ = γE ∗ (cos θ∗ +β) where γ = Eπ /mπ • Same as part (a), here we assume that the neutrinos are massless. So, in the pion rest frame, p∗ν = p∗ (1, sin θ∗ , 0, cos θ∗ ). Lorentz-transforming this vector to the lab frame, which is a boost in the z-direction with velocity −β, we have: pν = Λpν∗ ⎛ ⎞⎛ ⎞ γ 0 0 γβ 1 ⎜ 0 1 0 0 ⎟ ⎜ sin θ∗ ⎟ ⎟⎜ ⎟ = p∗ ⎜ ⎝ 0 0 1 0 ⎠⎝ 0 ⎠ γβ 0 0 γ cos θ∗ ⎛ ∗ ⎞ γp + γβ sin θ∗ p∗ ⎜ ⎟ cos θ∗ p∗ ⎟ = ⎜ ⎝ ⎠ 0 ∗ ∗ ∗ γβp + γ cos θ p Since pν is the momentum of the neutrino in the lab frame, we can also write it as pν = E(1, sin θ, 0, cos θ). Equating the two expression for the neutrino momentum, we immediate acquire: E = γE ∗ (1 + β cos θ∗ ) E cos θ = γE ∗ (cos θ∗ + β). Note that here p∗ ≡ E ∗ . (c) Using the expressions for E ∗ and θ∗ in terms of E and θ, show that γ 2 (1 − β cos θ)(1 + β cos θ∗ ) = 1. 2 (2) (3) • Dividing Equation 3 by Equation 2, we have: cos θ∗ + β 1 + β cos θ∗ 1 − β2 1 − β cos θ = 1 + β cos θ∗ γ −2 1 − β cos θ = 1 + β cos θ∗ (1 − β cos θ)(1 + β cos θ∗ )γ 2 = 1 cos θ = (4) (5) (d) Show that the maximum value of θ in the labortory frame is θmax = 1/γ. • θ is maximum when θ∗ = π2 ; i.e., the neutrino trajectory is perpendicular to the boost direction. Using Equation 4 and cos π2 = 0, we see that: cos θmax = β = p 1 − γ −2 . Since neutrinos are nearly massless, we can safely assume θmax ≈ 0 and γ  1. We than Taylor expand both sides of the equation above: 1− 2 θmax γ −2 = 1− 2 2 θmax = γ −1 (e) In the limit θ  1 show that E ≈ 0.43Eπ 1+βγ1 2 θ2 and therefore on-axis (θ = 0) the neutrino energy spectrum follows that of the pions. • Starting from Equation 2, and citing results from Equation 1 and 5, we have: 1 Eπ m2π − m2µ E= . mπ 2mπ (1 − β cos θ)γ 2 In the limit θ  1, we know that cos θ ≈ 1 − θ2 2 and that β ≈ 1. So, γ 2 θ2 β 2 1 γ 2 θ2 β = + 1+β 2 2 2 1 γ θ β ≈ + 2 2 (1 − β cos θ)γ 2 ≈ γ 2 (1 − β) + 3 (6) So, Equation 6 becomes: m2π − m2µ 1 2 mπ 1 + γ 2 θ2 β 1 E ≈ 0.43Eπ 1 + γ 2 θ2 β E = Eπ (7) (f) Assuming that the pions have a flat spectrum in the range 1-5 GeV, sketch the form of the resulting neutrino energy spectrum at the T2K far detector (SuperKamiokande), which is off-axis at θ = 2.5◦ . Given that the Super-Kamiokande detector is 295 km from the beam, explain why this angle was chosen. • By Equation 7, the neutrino energies at θ = 0◦ and θ = 2.5◦ for Eπ between 1-5 GeV are: So, the neutrino energy spectrum at θ = 2.5◦ is narrowly peaked, as opposed to the on-axis neutrinos which have a flat energy spectrum. This narrow peak maximizes the effect of neutrino oscillation at the T2K detector. Problem 2: Nuclear Stability [30 points] The Weizäcker formula or semi-empirical mass formula is a parametrization of nuclear mass as a function of A and Z. Following this formula, the mass of an atom with Z protons and N neutrons is given by the following: 2 −Z) M (A, Z) = N Mn + ZMp + Zme − aV A + as A2/3 + ac AZ1/3 + aa (N4A + with N = A − Z. 2 δ A1/2 The exact values of the parameters aV , as , ac , aa , and δ depend on the range of masses for which they are optimized. One possible set of parameters is given by the following: 4 aV = 15.67 MeV/c2 , as = 17.23 MeV/c2 , ac = 0.714 MeV/c2 , aa = 93.15 MeV/c2 and δ = −11.2, 0, +11.2 MeV/c2 for even Z and Z, odd A, or odd Z and N , respectively. For fixed A find the proton number Z for the most stable nucleus, and plot Z as a function of A. Each term captures an aspect of the atom. Explain briefly how the individual terms can be interpreted. • The most stable nucleus will have the highest binding energy, which is: EB (A, Z) = N Mn + ZMp + Zme − M (A, Z) (N − Z)2 δ Z2 − 1/2 = aV A − as A2/3 − ac 1/3 − aa 4A A A 2 = α + βZ + γZ . We take the derivative of EB to find the maximum: dEB (A, Z) = β + 2γZ = 0 dZ β aa A ⇒ Zstable = − = . 2γ 2(ac A2/3 + aa ) Note that if we assume the only type of decays here are conversion between nuclides with fixed A, then we should take derivative w.r.t. M (A, Z) instead of the binding energy. The dashed line is A = 2Z. 5 Interpretation of the terms: (also see pp. 20 of Particles and Nuclei: An Introduction to the Physical Concepts by Povh, Rith, Scholz and Zetsche, 5th Ed.) aV : the volume term. This term, which dominates the binding energy, is proportional to the number of nucleons. Each nucleon in the interior of a (large) nucleus contributes an energy of about 16 MeV. as : the surface term. For nucleons at the surface of the nucleus, which are surrounded by fewer nucleons, the above binding energy is reduced. This contribution is proportional to the surface area of the nucleus. ac : the Coulomb term. The electrical repulsive force acting between the protons in the nucleus further reduces the binding energy. aa : the asymmetry term. As long as mass numbers are small, nuclei tend to have the same number of protons and neutrons. Heavier nuclei accumulate more and more neutrons, to partly compensate for the increasing Coulomb repulsion by increasing the nuclear force. This creates an asymmetry in the number of neutrons and protons. The dependence of the nuclear force on the surplus of neutrons is described by the asymmetry term. δ: the pairing term. A systematic study of nuclear masses shows that nuclei are more stable when they have an even number of protons and/or neutrons. This observation is interpreted as a coupling of protons and neutrons in pairs. The pairing energy depends on the mass number, as the overlap of the wave functions of these nucleons is smaller, in larger nuclei. Problem 3: Decay time dating [20 points] Naturally occurring uranium is a mixture of the topes. 238 U (99.28%) and 235 U (0.72%) iso- How old must the material of the solar system be if one assumes that at its creation both isotopes were present in equal quantities? The lifetimes are τ (235 U) = 1 × 109 years and τ (238 U) = 6.6 × 109 years. • The number of exponentially-decaying nuclei as a function of time is: N (t) = N0 e−t/τ . Let τ1 = 6.6 × 109 yrs and τ2 = 1 × 109 yrs be the lifetime of 238 U and 235 U respectively, and let α be the ratio of abundance between 238 U and 235 U. Then, 6 α = N0 e−t/τ1 N0 e−t/τ2 α = e t = t = ( τ1 − τ1 )t 2 1 ln α − τ11 1 τ2 ln 99.28 0.72 1 1 − 9 6.6×109 yrs 1×10 yrs t = 5.8 × 109 yrs How much of the years ago? • 238 U has decayed since the formation of the earth’s crust 2.5 × 109 N = e−t/τ1 N0 9 N − 2.5×10 yrs = e 6.6×109 yrs N0 N ≈ 0.685. N0 So, about 31.5% of the 238 U as decayed. 7