複選題20．Select the correct electron configurations and quantum numbers considering LS coupling for various atoms. (A)Ca: 1s²2s²2p⁶3s²3p⁶4s²3d¹; 3d¹: n=3, ℓ=2, mℓ=0,±1,±2 (B)Cu: 1s²2s²2p⁶3s²3p⁶4s²3d⁹; 3d⁹: n=3, ℓ=2, mℓ=0,±1,±2 (C)Cr: 1s²2s²2p⁶3s²3p⁶4s¹3d⁵; 3d⁵: n=3, ℓ=2, mℓ=0,±1,±2, ms=+or- (D)Ca: 1s²2s²2p⁶3s²3p⁶4s²3d⁵; 3d⁵: n=3, ℓ=2, mℓ=0,±1,±2, ms=+or- (E)K: 1s²2s²2p⁶3s²3p⁶4s¹; 4s¹: n=4, ℓ=0, mℓ=0, ms=±1/2, mj=±

1. For a particle with an energy of E tunneling through a finite barrier height of U and width of & solve the Schrodinger wave equation to calculate the tunneling probability and select the correct answer corresponding to various E, U, and 8 for different particles. (A) For a neutron with E = 50 meV incident on a barrier of U = 250 meV and 8 = 0.16 nm, the tunneling probability is 4.45×10-11 = 0.0445 ppb (B) For a neutron with E = 50 meV incident on a barrier of U = 250 meV and 8 = 0.1 nm, the tunneling probability is 5.86×10-10 = 0.586 ppb (C) For an electron with E = 1.0 eV incident on a barrier with U = 5.0 eV and 8 = 0.5 nm, the tunneling probability is 7.1×10-7 = 0.71 ppm (D) For an electron with E = 1.0 eV incident on a barrier with U = 5.0 eV and 8 = 1.0 nm, the tunneling probability is 2.52×10-9 =2.52 ppb (E) For an electron with E = 1.0 eV incident on a barrier with U = 5.0 eV and & = 1.25 nm, the tunneling probability is 1.5×10-10 = 0.15 ppb

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