Exam (elaborations) TEST BANK FOR Introduction to Elementary Particles

Exam (elaborations) TEST BANK FOR Introduction to Elementary Particles For an undeflected charged particle, qE = qvB =⇒ v = E B . With just a magnetic field, qvB = m v 2 R =⇒ q m = v BR = E B2R . Problem 1.2 r0 = 10−15 m; ¯h = 6.58 × 10−22 MeV s; c = 3.00 × 108 m/s; so m = h¯ 2r0c =  hc¯ 2r0  1 c 2 = 98.7 MeV/c 2 . Observed mπ = 138 MeV/c 2 . Off by a factor of 1.4. Problem 1.3 r0 = 10−15 m; ¯h = 6.58 × 10−22 MeV s; c = 3.00 × 108 m/s; me = 0.511 MeV/c 2 . ∆x∆p ≥ h¯ 2 so pmin = h¯ 2r0 =  hc¯ 2r0  1 c = 98.7 MeV/c . Emin = q p 2 minc 2 + m2 e c 4 = 98.7 MeV . The energy of an electron emitted in the beta decay of tritium is 17 keV. 2 1 Historical Introduction to the Elementary Particles Problem 1.4 mΛ = 1 3 [2 (mN + mΞ) − mΣ] . mN = 938.9; mΞ = 1318.1; mΣ = 1190.5. So mΛ = 1 3 [2 (2257.0) − 1190.5] = 1107.8 MeV/c 2 . Observed mΛ = 1115.7 MeV/c 2 . Off by 0.7%. Problem 1.5 m 2 η = 1 3 h 2  m 2 K + m 2 K  − m 2 π i = 1 3  4m 2 K − m 2 π  . mK = 495.67; mπ = 138.04. m 2 η = 1 3 h 9.637 × 105 i = 3.212 × 105 ⇒ mη = 566.8 MeV/c 2 . Actually mη = 547.3 MeV/c 2 . Off by 3.5%. Problem 1.6 M∆ − MΣ∗ = 1232 − 1385 = −153 MΣ∗ − MΞ∗ = 1385 − 1533 = −148. Average: −151. ∴ MΩ = MΞ∗ + 151 = 1533 + 151 = 1684 MeV/c 2 . Actually MΩ = 1672 MeV/c 2 . Off by 0.7%. Problem 1.7 (a) ∆ − −→ n + π − or Σ − + K 0 Σ ∗+ −→ p + ¯K0 ; Σ + + π 0 ; Σ + + η; Σ 0 + π +; Λ + π +; Ξ 0 + K + Ξ ∗− −→ Σ 0 + K −; Σ − + ¯K0 ; Λ + K −; Ξ 0 + π −; Ξ − + π 0 ; Ξ − + η 3 (b) Kinematically allowed: ∆ − −→ n + π − Σ ∗+ −→ Σ + + π 0 ; Σ 0 + π +; Λ + π + Ξ ∗− −→ Ξ 0 + π −; Ξ − + π 0 Problem 1.8 (a) With a strangeness of −3, the Ω− would have to go to