Summary Flashcards

Question

Eigenvalues of L̂ᶻ

Answer 1

l and m*ₗ* are integers, with l ≥ 0 and −l ≤ m*ₗ* ≤ l

Answer 2

Spherical harmonics, Yₗₘₗ(θ, ϕ).

Answer 3

L̂ᶻ = −iħ ∂/∂ϕ L̂² = −ħ² (∂²/∂θ² + cotθ ∂/∂θ + 1/sin²θ ∂²/∂ϕ²) | L̂² only needs to be recognised.

Answer 4

Ĥ = −ħ²/2mₑ (d²/dr² + 2/r d/dr) + L̂²/2mₑr² − e²/4πε₀r | Bracket only needs to be recognised

Answer 5

Eₙ = −Eᴿ/n² where Eᴿ = mₑe⁴/32π²ε₀²ħ² = 13.6 eV | mₑ term only needs to be recognised, ᴿ should be subscript

Answer 6

[Ĥ, L̂²] = 0 and [Ĥ, L̂ᶻ] = 0 Measurement of each quantum number doesnt affect the others.

Answer 7

ψₙ,ₗ,ₘₗ = const (r/a₀)*ˡ* exp(−r/na₀) × Qₙ,ₗ (r/a₀) × Pₗ,ₘₗ (θ)exp(im*ₗ*ϕ)

Answer 8

Set is called the n“l” orbital. “l” = s, p, d, f . . . for l = 0, 1, 2, 3, . . ..

Answer 9

Eᴿ → Z²Eᴿ a₀ → a₀/Z. | ᴿ should be subscript

Answer 10

If Ĥ = Ĥ₀ + V̂ and V̂ is small, use eigenfunctions of Ĥ₀ with energy shift ⟨V̂⟩

Answer 11

⟨Ŝ²⟩ = s(s + 1)ħ² s = ½ ⟨Ŝᶻ⟩ = mₛħ mₛ = ±½

Answer 12

µ̂ = −µᴮ 1/ħ (L̂ + gŜ) where µᴮ = eħ/2mₑ and g ≈ 2

Answer 13

Separates a beam of atoms into different components, depending on their µᶻ.

Answer 14

Ĵ = L̂ + Ŝ ⟨Ĵ²⟩ = j(j + 1)ħ² ⟨Ĵᶻ⟩ = mⱼħ mⱼ between −j and j j can be in integer steps between |l − s| and l + s. ⟨L̂²⟩ =l(l + 1)ħ² ⟨L̂ᶻ⟩ = mₗħ mₗ between −l and l l is integer 0,1,2... ⟨Ŝ²⟩ = s(s + 1)ħ² ⟨Ŝᶻ⟩ = mₛħ mₛ = ±½ s = ½

Answer 15

V̂ ∝ L̂ · Ŝ = ½ (Ĵ² − L̂² − Ŝ²)

Answer 16

n - principal quantum number - linked to energy level - 1, 2, 3... l - azimuthal quantum number - orbital angular momentum - s,p,d,f correspond to 0,1,2,3 - determines total electrons possible in each subshell 2,6,10,14 - possible values 0 to n-1 j - main total angular momentum quantum number - integer steps between |l - s| and l+s (s is spin quantum number) mⱼ - secondary total angular momentum quantum number - integer steps between -j and j

Answer 17

If field “weak”, ⟨V̂⟩ = g*ₗ*µᴮBmⱼ where g*ₗ* = “Landé g factor” if field “strong”, ⟨V̂⟩ = µᴮB(m*ₗ* + gmₛ) “Weak” and “strong” are relative to size of spin-orbit energy. | g*ₗ* should be L, µᴮ should be subscript B

Answer 18

x → −x

Answer 19

Only allowed between states of different parity.

Answer 20

Fundamental particles are indistinguishable from each other, meaning it is impossible to tell if they have swapped places.

Answer 21

{x1, s1} ↔ {x2, s2} Wavefunctions must be symmetric or antisymmetric under exchange.

Answer 22

Particles being indistinguishable means that |ψ(x₁, x₂)|² = |ψ(x₂, x₁)|² This is only possible if ψ(x₁, x₂) = ψ(x₂, x₁) OR ψ(x₁, x₂) = -|ψ(x₂, x₁) for ψ(x₁, x₂) = ψ(x₂, x₁), the particles are symmetric, whereas for ψ(x₁, x₂) = -ψ(x₂, x₁), they are anti-symmetric

Answer 23

Symmetric: bosons, such as photons Antisymmetric: fermions, such as electrons

Answer 24

Electron wavefunctions have either Symmetric spin | antisymmetric space or Symmetric space | antisymmetric spin

Answer 25

Two electrons cannot be in an identical state

Answer 26

Helium wavefunctions are approximately the product of two hydrogen wavefunctions plus an upward energy shift due to energy of Coulomb repulsion.

Answer 27

(1s)² 2 electrons in 1s orbital with antisymmetric electron spins.

Answer 28

Exchange energy is a shift in repulsion energy due to fact that electrons are more often closer together/further apart in symmetric/antisymmetric combinations than in independent wavefunctions. It links to parallel and identical electron spin.

Answer 29

* Orbitals fill from lowest energy to highest * Lower orbitals shield nuclear charge from higher orbitals * Lower l has lower energy because electron spends more time close to origin, inside inner shell and ‘sees’ more of the nucleus. * Filled orbitals have zero total angular momentum and zero total spin.

Answer 30

* Maximize S (fill all m*ₗ* ‘slots’ with aligned spins before any with anti-aligned) * Maximize L (fill slots from high m*ₗ* to low, with L = Σm*ₗ*) * If orbital is less than half full, minimize J = |L − S|. * If orbital is more than half full, maximize J = L + S.

Answer 31

²ˢ⁺¹“L”ⱼ e.g. Fe = ⁵D₄

Answer 32

1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 6s 6p 7s Order diagonally top right to bottom left 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, . . .