14. What is the probability of finding a n = 2, /= 1 electron 26 between 15 a For a hydrogen and 2ao? Orbital angular momentum Orbital magnetic quantum number |L|= √√(+1)h 7.2 Angular probability P(0, 4) = 7.5 (1= 0, 1, 2,...) density |0lm,(0)m, ($)|2 L₁ = m₁h 7.2 Orbital magnetic ₁ =-(e/2m)L 7.6 (m,=0,11,12,..., ±1) dipole moment Spatial quantization L₁₂ m₁ cos 0 = = 7.2 Spin magnetic dipole s = -(e/m)Š 7.6- |L| √(+1) moment Angular momentum uncertainty relationship Hydrogen quantum numbers Hydrogen energy levels Hydrogen wave functions ΔΙΑΦ ΣΑ n = 1,2,3,... = 0, 1, 2,...,n-1 m₁ =0,11,12,...,±l 7.2 Spin angular |S|= √√s(s+1)h= 7.6 momentum √√3/4h (for s = 1/2) Spin magnetic 7.3 S=mh (m,±2) 7.6 quantum number Spectroscopic notation En=- me 1 32л²² n² 7.3 7.3 Selection rules for photon emission Normal Zeeman effect s(10), p (1 = 1), d(l=2),f(l= 3),... ΔΙ = +1 Δ = 0,11 7.7 7.7, 7.8 2² R()(0) (0) Radial probability density P(r) = r²|R(r)|2 7.4 Fine-structure estimate Δλ= AE = mc2a/n = AE = BB 7.8 7.9 (a≈1/137)

14. What is the probability of finding a n = 2, /= 1 electron 26 between 15 a For a hydrogen and 2ao? Orbital angular momentum Orbital magnetic quantum number |L|= √√(+1)h 7.2 Angular probability P(0, 4) = 7.5 (1= 0, 1, 2,...) density |0lm,(0)m, ($)|2 L₁ = m₁h 7.2 Orbital magnetic ₁ =-(e/2m)L 7.6 (m,=0,11,12,..., ±1) dipole moment Spatial quantization L₁₂ m₁ cos 0 = = 7.2 Spin magnetic dipole s = -(e/m)Š 7.6- |L| √(+1) moment Angular momentum uncertainty relationship Hydrogen quantum numbers Hydrogen energy levels Hydrogen wave functions ΔΙΑΦ ΣΑ n = 1,2,3,... = 0, 1, 2,...,n-1 m₁ =0,11,12,...,±l 7.2 Spin angular |S|= √√s(s+1)h= 7.6 momentum √√3/4h (for s = 1/2) Spin magnetic 7.3 S=mh (m,±2) 7.6 quantum number Spectroscopic notation En=- me 1 32л²² n² 7.3 7.3 Selection rules for photon emission Normal Zeeman effect s(10), p (1 = 1), d(l=2),f(l= 3),... ΔΙ = +1 Δ = 0,11 7.7 7.7, 7.8 2² R()(0) (0) Radial probability density P(r) = r²|R(r)|2 7.4 Fine-structure estimate Δλ= AE = mc2a/n = AE = BB 7.8 7.9 (a≈1/137)