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

```THE HYDROGEN ATOM
CH 7: SECTIONS 6 AND 7
homework
• Ch 7: 37
• (Due 10DEC,
Thursday—same day
as Test 4)
β2 1 ππ 2 ππ
β β+1
2
−
ππ
ππ
ππ
+
β
ππ ππ + ππ ππ ππ ππ = πΈπΈπΈπΈ(ππ)
2
2
ππππ
2ππππ
2ππ ππ ππππ
• Statement of conservation of energy.
• Now to specifically consider hydrogen atom, substitute ππ(ππ) =
and solve the equation.
• We’ve already seen quantized energy:
πΈπΈππ =
ππ2ππππ4 1
− 2 2
2β ππ
= −
ππππ2
−
ππ
13.6 eV
ππ2
• And quantized radii from Bohr’s model (not quite right): ππ = ππ0 ππ2
• Where n (principle quantum number) = 1, 2, 3, …
• And the Bohr radius ππ0 =
β2
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: πΈπΈππ =
ππ2ππππ4 1
− 2 2
2β ππ
= −
13.6 eV
ππ2
Principle Quantum Number: n = 1, 2, 3 …..
• Angular momentum: πΏπΏ =
β(β + 1)β
Orbital Quantum Number: β = 0, 1, 2, … (n − 1)
• Z-component of angular momentum: πΏπΏπ§π§ = ππβ β
Magnetic quantum number: ππβ = 0, &plusmn;1, &plusmn;2 … &plusmn; β
Spherical and
Solutions
Notice the Bohr radius in
ππ0 = 0.0529 nm
Degeneracy
Spherical symmetry
Electron
probability
density
Labeled by
spectroscopic notation
Another way to show probability density