THE HYDROGEN ATOM RADIAL PROBABILITIES CH 7: SECTIONS 6 AND 7 homework • Ch 7: 37 • (Due 10DEC, Thursday—same day as Test 4) RADIAL DIFF.EQ. β2 1 ππ 2 ππ β β+1 2 − ππ π π ππ + β π π ππ + ππ ππ π π ππ = πΈπΈπΈπΈ(ππ) 2 2 ππππ 2ππππ 2ππ ππ ππππ • Statement of conservation of energy. • Now to specifically consider hydrogen atom, substitute ππ(ππ) = and solve the equation. • We’ve already seen quantized energy: πΈπΈππ = ππ2ππππ4 1 − 2 2 2β ππ = − ππππ2 − ππ 13.6 eV ππ2 • And quantized radii from Bohr’s model (not quite right): ππ = ππ0 ππ2 • Where n (principle quantum number) = 1, 2, 3, … • And the Bohr radius ππ0 = β2 ππππππ2 = 0.0529 nm • And the solution also tells us about the orbital quantum number’s limit: β = 0, 1, 2, … (n − 1) Hydrogen quantum numbers and quantized properties (memorize) • Energy: πΈπΈππ = ππ2ππππ4 1 − 2 2 2β ππ = − 13.6 eV ππ2 Principle Quantum Number: n = 1, 2, 3 ….. • Angular momentum: πΏπΏ = β(β + 1)β Orbital Quantum Number: β = 0, 1, 2, … (n − 1) • Z-component of angular momentum: πΏπΏπ§π§ = ππβ β Magnetic quantum number: ππβ = 0, ±1, ±2 … ± β Spherical and Radial Solutions Notice the Bohr radius in the radial solutions: ππ0 = 0.0529 nm Degeneracy Spherical symmetry leads to degeneracy Electron probability density Labeled by spectroscopic notation Another way to show probability density Radial Probability • Recall ππππ = ππ2 sin ππ ππππππππππππ • RADIAL Probability per unit radial distance: ππ = π π 2ππ2 • The r2 is because it is more likely to find particle in thin large sphere than in thin small sphere, and this depends on the surface area: Radial Probability In class problems • Ch 7: 44, 45, 56 • Know how to do these for Test 4