Quantum Physics Lecture 10

-  Solution of Schroedinger equation -  3 quantum numbers -  Probability and ‘Orbitals’ -  Selection rules for transitions -  Zeeman Effects, electron spin

The Hydrogen atom

Many electron atoms (Z>1)

- Periodic table of elements - Moseley’s Law

Hydrogen Atom Re-visited First real test of Schroedinger Equation:

Recall Bohr model: En ∝ -1/n2, via one quantum number n (also, quantised angular momentum L via same n)

Successes: Hydrogen line emission spectra also… size & stability of atoms, Moseley’s Law for Z, etc.

Problems: Full 3D & spherical symmetry? Magnetic effects in spectra Z > 1 atoms’ spectra

Comparison with Bohr model: same expression for En via “principal” quantum number n

L = n

Schroedinger Equation for H atom

Ψ r,θ ,φ( ) = R r( )Θ θ( )Φ φ( )

Express wavefunction as product (c.f 2-D box)

Leads to 3 separate equations

−

2

2mr 2

ddr

r 2dR r( )

dr

⎛

⎝⎜

⎞

⎠⎟ +

2l l +1( )2mr 2 +U r( )

⎛

⎝⎜

⎞

⎠⎟ R r( ) = ER r( )

1sinθ

ddθ

sinθdΘ θ( )

dθ

⎛

⎝⎜

⎞

⎠⎟ + l l +1( ) − ml

2

sin2θ

⎛

⎝⎜

⎞

⎠⎟ Θ θ( ) = 0

d 2Φ φ( )dφ 2 + ml

2Φ φ( ) = 0

R solutions are e-αr (× r polynomial) type, Θ are functions of cos & sin θ Φ solutions are exp(imlϕ) type, where ml is an integer

Solutions specified by three quantum numbers: n, l, ml

Schroedinger Equation for H atom (cont.)

Angular momentum - is quantised Solution shows

(c.f. angular momentum in lecture 9)

Also Φ must be single valued and

Find

Note that L can be zero (unlike Bohr model) And Lz < L (unless L=0)

∵ Exists an Uncertainty relation between Lx Ly Lz

Bohr – plane, but must have Δp normal to plane… Also seen via (try it! – commutator)

But so no uncertainty between L and Lz & can know both

Φ φ( ) = Φ φ + 2π( )

Lz = ml (ml = 0,±1,± 2... ± l)

L = l l +1( )

‘Semi-classical’ vector pictures for l=2

Lx Ly − Ly Lx = Lx , Ly

⎡⎣ ⎤⎦ = iLz

L2 , Lz⎡⎣ ⎤⎦ = 0

Solutions specified by three quantum numbers: n, l, ml (c.f. 3-D box, Lect. 8)

Energy Levels are En and the same as the Bohr model!

with n = 1, 2, 3, 4, ….

Solutions show l = 0, 1, 2, 3… n-1 and | ml | ≤ l Notation: s, p, d, f… for l = 0, 1, 2, 3, ….

So, possible states of electron:

Schroedinger Equation for H atom (cont.)

En = −me4

8εo2h2

1n2( )

Note: En does not depend on l, ml i.e. l, ml states are degenerate –

but holds only for H atom

Electron states in hydrogen atom - orbitals

s-states - spherically symmetric

P r( )dr = ψ

24πr 2dr

Radial probability distributions (a = Bohr radius)

Note for l = n-1 (1s, 2p, 3d…) there is one maximum, at r = n2a

The (ml) sub-shells combine to give spherically symmetric (l) shells - Spherical ‘harmonics’ (complete shells…)

e.g. for 2p-states, cosθ & sinθ dependence Squared (probability) & add - gives 1

Angular functions - orbitals

2p orbitals

Images of orbitals….

N.B. Images (e.g. from web) are often of probability |ψ|2 rather than ψ

Recall that ψ can have negative as well as positive values. For example one lobe of the 2px orbital is +ve the other lobe -ve. Both give the same probability

Transition of electrons between orbitals

Selection rules for transitions: Δl = ± 1 And Δml = 0 or ± 1 All others are ‘forbidden’

Conservation overall of angular momentum, photon has angular momentum of

l determines the magnitude of L, ml determines the direction

A non-zero magnetic field direction defines z But B field alters the energy... N.B. Magnetic dipole moment (current in loop x area of loop)

with (Bohr Magneton) e+

e- µ =

ve2πr

πr 2 mvr =

µB =

e2m

Applied B field shifts energy (dipole-field interaction) so ml states are split apart by µBB

So moment = mlµB

Zeeman Effect

In non-zero magnetic field B each l-level splits into 2l+1 ml sub-levels of slightly different energy, depending on B

Spectra show three lines instead of one; only three because of further selection rule:

Δml = ± 1, 0

But…. some transitions showed more than three - anomalous

Anomalous Zeeman Effect

Problem: more than 3 lines and smaller separation(s)

Solution: “electron spin”

See lecture 9 !

Electron has spin angular momentum but

Spin angular momentum S (cf L) Spin magnetic quantum number ms (cf ml)

Total combined angular momentum J

Combining L and S leads to correct number and splitting of emission lines (first seen by Irishman Thomas Preston)

J =L +S

J = j j +1( )

j = l ± 1

2

sz = ±

2 µz = − 2.000232( ) e

2msz

More than 1 electron atoms…

Simple picture:

can add electrons to hydrogen-like quantum states assumes the nucleus increasing charge is shielded by inner electrons and ignores electron – electron interaction energies…. Try it anyway….

But: Pauli exclusion principle – no two electrons can occupy the same quantum state (fermions, see Lecture 9)

Two spins states possible – ‘up’ and ‘down’ ↑ or ↓

So maximum of two electrons in each state one ↑ and the other↓

±

2

Many electron atoms

For Z>1, fill quantum states with max. 2 electrons each First quantum number n (c.f.Bohr energy level) Second quantum number l (angular momentum state) where 0 ≤ l ≤ (n-1) Third quantum number ml where |ml| ≤ l

Many electron atoms (cont.) Examples: Helium atom Contains 2 electrons, both can be in 1s state (lowest energy)

provided one is spin up the other spin down Notation for the ground state 1s2

Lithium (Z=3)

1s shell filled (like He) Extra electron goes into 2s shell

Notation 1s2 2s1

2s orbital further out… Nuclear charge screened by 1s shell, effective charge more like H but further out So less well bound 2s electron can be lost in bonding (ionicity)

n=1 - l=0 s state – 2 electrons n=2 - l=1 p state – 6 electrons …… l=2 d state – 10 electrons ……. l=3 f state – 14 electrons

Z>1 The periodic table of elements Gives the basic structure of the Periodic Table of the elements

Periodic Table of Elements

For Hydrogen, s,p,d,f, states have same energy for given n (c.f. Bohr)

This Degeneracy of states is broken for Z >1 (by e-e interaction potentials) So s fills before p, before d etc., the gap increasing

as Z becomes larger.

1-7s 4-5f 3-6d 2-7p

X-Ray spectra

X-Rays are emitted by impact of high energy electrons on elements

Continuous spectrum due Bremsstrahlung & other scattering processes

Molybdenum spectrum shown

λmin =

hceV

Impacting electrons cause electrons in core (lowest energy) states to be knocked out. For high Z atoms, these are very tightly bound states (K shells), so require high energies (many keV) to eject them

Spectrum shows sharp peaks, due to emission of photons by outer electrons falling to vacated core states. Energy (frequency) is characteristic of element.

N.B. Lower energy spectroscopy shows energies which often have little to do with the Z number of the atom – a problem for early atom models!

Moseley’s Law

Moseley found that

The first time Z was spectroscopically determined…

f ∝ Z −1( )2

One other electron in K-shell, so nuclear charge screened by 1e, i.e. reduced to Z-1 Transition from n=2 to n=1 gives (Bohr model)

Which agrees very closely with Moseley’s experiment.

Actually the most important early evidence for nuclear model of atom!

ΔE =

34

Ryd Z −1( )2