THE HYDROGEN ATOM RADIAL PROBABILITIES CH 7: SECTIONS 6 AND 7

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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.
ℏ2 1 𝑑𝑑 2 𝑑𝑑
β„“ β„“+1
2
−
π‘Ÿπ‘Ÿ
𝑅𝑅
π‘Ÿπ‘Ÿ
+
ℏ
𝑅𝑅 π‘Ÿπ‘Ÿ + π‘ˆπ‘ˆ π‘Ÿπ‘Ÿ 𝑅𝑅 π‘Ÿπ‘Ÿ = 𝐸𝐸𝐸𝐸(π‘Ÿπ‘Ÿ)
2
2
𝑑𝑑𝑑𝑑
2π‘šπ‘šπ‘šπ‘š
2π‘šπ‘š π‘Ÿπ‘Ÿ 𝑑𝑑𝑑𝑑
• Statement of conservation of energy.
• Now to specifically consider hydrogen atom, substitute π‘ˆπ‘ˆ(π‘Ÿπ‘Ÿ) =
and solve the equation.
• We’ve already seen quantized energy:
𝐸𝐸𝑛𝑛 =
π‘˜π‘˜2π‘šπ‘šπ‘šπ‘š4 1
− 2 2
2ℏ 𝑛𝑛
= −
π‘˜π‘˜π‘˜π‘˜2
−
π‘Ÿπ‘Ÿ
13.6 eV
𝑛𝑛2
• And quantized radii from Bohr’s model (not quite right): π‘Ÿπ‘Ÿ = π‘Žπ‘Ž0 𝑛𝑛2
• Where n (principle quantum number) = 1, 2, 3, …
• And the Bohr radius π‘Žπ‘Ž0 =
ℏ2
π‘˜π‘˜π‘˜π‘˜π‘˜π‘˜2
= 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: 𝐸𝐸𝑛𝑛 =
π‘˜π‘˜2π‘šπ‘šπ‘šπ‘š4 1
− 2 2
2ℏ 𝑛𝑛
= −
13.6 eV
𝑛𝑛2
Principle Quantum Number: n = 1, 2, 3 …..
• Angular momentum: 𝐿𝐿 =
β„“(β„“ + 1)ℏ
Orbital Quantum Number: β„“ = 0, 1, 2, … (n − 1)
• Z-component of angular momentum: 𝐿𝐿𝑧𝑧 = π‘šπ‘šβ„“ ℏ
Magnetic quantum number: π‘šπ‘šβ„“ = 0, ±1, ±2 … ± β„“
Spherical and
Radial
Solutions
Notice the Bohr radius in
the radial solutions:
π‘Žπ‘Ž0 = 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 𝑑𝑑𝑑𝑑 = π‘Ÿπ‘Ÿ2 sin πœƒπœƒ π‘‘π‘‘π‘‘π‘‘π‘‘π‘‘πœƒπœƒπ‘‘π‘‘πœ™πœ™
• RADIAL Probability per unit radial distance: 𝑃𝑃 = 𝑅𝑅2π‘Ÿπ‘Ÿ2
• 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
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