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
ℓ ℓ+1
2
−
𝑟𝑟
𝑅𝑅
𝑟𝑟
+
ℏ
𝑅𝑅 𝑟𝑟 + 𝑈𝑈 𝑟𝑟 𝑅𝑅 𝑟𝑟 = 𝐸𝐸𝐸𝐸(𝑟𝑟)
2
2
𝑑𝑑𝑑𝑑
2𝑚𝑚𝑚𝑚
2𝑚𝑚 𝑟𝑟 𝑑𝑑𝑑𝑑
• Statement of conservation of energy.
• Now to specifically consider hydrogen atom, substitute 𝑈𝑈(𝑟𝑟) =
and solve the equation.
• We’ve already seen quantized energy:
𝐸𝐸𝑛𝑛 =
𝑘𝑘2𝑚𝑚𝑚𝑚4 1
− 2 2
2ℏ 𝑛𝑛
= −
𝑘𝑘𝑘𝑘2
−
𝑟𝑟
13.6 eV
𝑛𝑛2
• And quantized radii from Bohr’s model (not quite right): 𝑟𝑟 = 𝑎𝑎0 𝑛𝑛2
• Where n (principle quantum number) = 1, 2, 3, …
• And the Bohr radius 𝑎𝑎0 =
ℏ2
𝑘𝑘𝑘𝑘𝑘𝑘2
= 0.0529 nm
• And the solution also tells us about the orbital quantum number’s
limit: ℓ = 0, 1, 2, … (n − 1)
Hydrogen quantum numbers and
quantized properties (memorize)
• Energy: 𝐸𝐸𝑛𝑛 =
𝑘𝑘2𝑚𝑚𝑚𝑚4 1
− 2 2
2ℏ 𝑛𝑛
= −
13.6 eV
𝑛𝑛2
Principle Quantum Number: n = 1, 2, 3 …..
• Angular momentum: 𝐿𝐿 =
ℓ(ℓ + 1)ℏ
Orbital Quantum Number: ℓ = 0, 1, 2, … (n − 1)
• Z-component of angular momentum: 𝐿𝐿𝑧𝑧 = 𝑚𝑚ℓ ℏ
Magnetic quantum number: 𝑚𝑚ℓ = 0, ±1, ±2 … ± ℓ
Spherical and
Radial
Solutions
Notice the Bohr radius in
the radial solutions:
𝑎𝑎0 = 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 = 𝑟𝑟2 sin 𝜃𝜃 𝑑𝑑𝑑𝑑𝑑𝑑𝜃𝜃𝑑𝑑𝜙𝜙
• RADIAL Probability per unit radial distance: 𝑃𝑃 = 𝑅𝑅2𝑟𝑟2
• 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