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Phys 102 – Lecture 26 The numbers and

1 Recall: the

Only orbits that fit n e– wavelengths are allowed

SUCCESSES Correct energy & atomic spectra mk24 e 1 E = −⋅e n =1231,2,3,... n 222n + FAILURES Radius & momentum quantization violates Heisenberg n22 2 Δr ⋅Δp ≥ rnan = 2 ≡ 0 r mkee 2 – e wave orbits cannot have zero L

Lnn = Orbits can hold any number of Phys. 102, Lecture 26, Slide 2 Quantum Mechanical

Schrödinger’s equation determines e– “wavefunction”

⎛⎞22ke 3 quantum numbers −∇−2 ψ(,r θφ , ) =Eψ (,r θφ , ) ⇒ ψ determine e– state ⎜⎟ nm,, ⎝⎠2mre “SHELL” “Principal ” n = 1, 2, 3, … mk24 e 1 E =− e Energy n 222n s, p, d, f “SUBSHELL” “Orbital Quantum Number” ℓ = 0, 1, 2, 3 …, n–1

L =+(1) Magnitude of

“” mℓ = –ℓ, …–1, 0, +1…, ℓ

Lz = m Orientation of angular momentum

Phys. 102, Lecture 26, Slide 3 ACT: CheckPoint 3.1 & more

For which state is the angular momentum required to be 0? A. n = 3 B. n = 2 C. n = 1

How many values for mℓ are possible for the f subshell (ℓ = 3)? A. 3 B. 5 C. 7

Phys. 102, Lecture 26, Slide 4 electron orbitals

2 ψ nm,, ∝ probability

Shell (n, ℓ, mℓ) (2, 0, 0) (2, 1, 0) (2, 1, 1) Subshell

(3, 0, 0) (3, 1, 0) (3, 1, 1) (3, 2, 0) (3, 2, 1) (3, 2, 2)

(4, 0, 0) (4, 1, 0) (4, 1, 1) (4, 2, 0) (4, 2, 1) (4, 2, 2)

(5, 0, 0) (4, 3, 0) (4, 3, 1) (4, 3, 2) (4, 3, 3) (15, 4, 0)

Phys. 102, Lecture 26, Slide 5 CheckPoint 2: orbitals

Orbitals represent probability of electron 1s (ℓ = 0) being at particular location r

2p (ℓ = 1) r

2s (ℓ = 0) r

3s (ℓ = 0)

r

Phys. 102, Lecture 26, Slide 6 Angular momentum

What do the quantum numbers ℓ and mℓ represent? Magnitude of angular momentum vector quantized + r L ==L (1) + = 0,1,2n − 1 Only one component of L quantized – – e Lz = m m =−,1,0,1, − Classical orbit picture Other components L , L are not quantized L y Lz = +2 L =+ L = 2 L = 6 z Lz = + L = 0 =1 z = 2 Lz = 0 L =− z Lz = −

Phys. 102, Lecture 26, Slide 7 Lz = −2 Orbital magnetic dipole

Electron orbit is a current loop and a magnetic dipole μ e e μe = IA =− L Recall Lect. 12 2me + r e μe =− L Dipole moment is quantized – 2m e– e L What happens in a B field? e B μ U = −μeB cosθ = Bm Recall Lect. 11 e 2m z θ e Orbitals with different L have different quantized energies in a B field – + e e – r −5 eV μB ≡ =×5.8 10 “Bohr magneton” 2me T Phys. 102, Lecture 26, Slide 8 ACT: dipole

What is the magnetic dipole moment of hydrogen in its ground stttate due to the orbita l of eltlectrons ?

e μe = − L 2me

e A. μH = − 2me

B. μH = 0 e C. μH =+ 2me

Phys. 102, Lecture 26, Slide 9 Calculation:

B Calculate the effect of a 1 T B field on the energy of the 2p (n = 2, ℓ = 1) level μe EEtot= n=2 − μ eB cosθ z m = −1 e = EBm+ For ℓ = 1, n=2 2m m = –1, 0, +1 + e ℓ B = 0 B > 0 m = +1 e– – m = 0

m = −1 splits into 3, with energy splitting eB Δ≡E 2me Phys. 102, Lecture 26, Slide 10 ACT: Atomic dipole

The H α spectral line is due to e– transition between the n = 3, ℓ = 2 and the n = 2, ℓ = 1 subshells.

How many spectltral lines will appear in a B fie ld? E A. 1 B. 3 ℓ = 2 C. 5 n = 3

n = 2 ℓ = 1 Phys. 102, Lecture 26, Slide 11 Intrinsic angular momentum

A beam of H in passes through a B field n = 1, so ℓ = 0 and expect NO effect from B field Atom with ℓ = 0

Instead, observe beam split in two!

Since we expect 2ℓ + 1 values for magnetic dipole moment, e– must have intrinsic angular momentum ℓ = ½. “Spin”

“Stern‐Gerlach experiment”

Phys. 102, Lecture 26, Slide 12 Spin angular momentum

Electrons have an intrinsic angular momentum called “spin” 3 S = + S = S ==Sss(1) + with s = ½ z 2 2

11 S = m ms =+= − ,+ zs 22 S = − z 2 Spin also generates magnetic dipole moment S e with g ≈ 2 B μs =− gS 2me S Spin DOWN (– ½) Spin UP (+½) ge 1 U = −μs B cosθ = Bms ms = + 2 2me B > 0 1 Phys. 102, Lecture 26, Slide 13 ms = − 2 Magnetic resonance

e– in B field absorbs photon with energy equal to splitting of energy levels

Absorption 1 “Electron spin resonance” ms = + 2 hf= Δ E Typically microwave EM wave 1 ms = − 2

Protons & also have spin ½ e μprot=+ g p S << μs since mmp >> e 2mp “Nuclear magnetic resonance”

Phys. 102, Lecture 26, Slide 14 Quantum number summary

“”, n = 1, 2, 3, … mk24 e 1 E =− e Energy n 222n “Orbital Quantum Number”, ℓ = 0, 1, 2, …, n–1 L =+(1) Magnitude of angular momentum

“Magnetic Quantum Number”, ml = –ℓ, … –1, 0, +1 …, ℓ

Lz = m Orientation of angular momentum

“”, ms = –½, +½

Smzs= Orientation of spin Electronic states

Pauli Exclusion Principle: no two e– can have the same set of quantum numbers E

s (ℓ = 0) p (ℓ = 1)

+½ –½ +½ –½ +½ –½ +½ –½

n = 2 – – – – – – – –

mℓ = –1 mℓ = 0 mℓ = +1

mS = +½ mS = –½

n = 1 – – Phys. 102, Lecture 26, Slide 16 The

Pauli exclusion & energies determine sequence s (ℓ = 0) ( ) Also s p (ℓ = 1) n = 1 2 d (ℓ = 2) 3 4 5 6 7

f (ℓ = 3) Phys. 102, Lecture 26, Slide 17 CheckPoint 3.2

How many electrons can there be in a 5g (n = 5, ℓ = 4) sub‐ shell of an atom?

Phys. 102, Lecture 26, Slide 18 ACT: Quantum numbers

How many total electron states exist with n = 2?

A. 2 B. 4 C. 8

Phys. 102, Lecture 26, Slide 19 ACT: Magnetic elements

Where would you expect the most magnetic elements to be in the periodic table?

A. Alkali metals (s, ℓ = 1) B. Noble gases (p, ℓ = 2) C. Rare earth metals (f, ℓ = 4)

Phys. 102, Lecture 26, Slide 20 Summary of today’s lecture

• Quantum numbers PiPrinc ipa l quantum number EVE = −13. 6eV n2 Orbital quantum number L = (1),+=− 0,1,1n

Magnetic quantum number Lmzz==−,,, m ……0, • Spin angular momentum – 11 e has intrinsic angular momentum Smzs= m s=−22, • Magnetic properties Orbital & spin angular momentum generate magnetic dipole moment • Pauli Exclusion Principle No two e– can have the same quantum numbers

Phys. 102, Lecture 25, Slide 21