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Quantum Physics Lecture 10
The Hydrogen atom
-  Solution of Schroedinger equation
-  3 quantum numbers
-  Probability and ‘Orbitals’
-  Selection rules for transitions
-  Zeeman Effects, electron spin
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)
L = 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
Schroedinger Equation for H atom
Express wavefunction as product (c.f 2-D box)
(
)
() () ()
Ψ r,θ ,φ = R r Θ θ Φ φ
Leads to 3 separate equations
()
(
)
2
⎞
 2 d ⎛ 2 dR r ⎞ ⎛  l l + 1
−
+ U r ⎟ R r = ER r
⎜r
⎟ +⎜
dr ⎠ ⎝ 2mr 2
2mr 2 dr ⎝
⎠
()
() ()
()
dΘ θ ⎞ ⎛
ml2 ⎞
1 d ⎛
⎜ sin θ
⎟ + l l +1 − 2 ⎟ Θ θ = 0
sin θ dθ ⎝
dθ ⎠ ⎜⎝
sin θ ⎠
(
)
()
( )+m Φ φ =0
(
)
dφ
d 2Φ φ
2
2
l
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
(
)
L = l l +1 
(c.f. angular momentum in lecture 9)
Also Φ must be single valued and Φ (φ ) = Φ (φ + 2π )
Find
Lz = ml  (ml = 0, ± 1, ± 2... ± l)
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 Lx Ly − Ly Lx = ⎡⎣ Lx , Ly ⎤⎦ = iLz
But
⎡ L2 , Lz ⎤ = 0
⎣
⎦
(try it! – commutator)
so no uncertainty between L and Lz & can know both
‘Semi-classical’ vector
pictures for l=2
Schroedinger Equation for H atom (cont.)
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!
( )
me 4 1
En = − 2 2
n2
8ε o h
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: 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
()
2
P r dr = ψ 4π r 2 dr
s-states - spherically symmetric
Radial probability distributions
(a = Bohr radius)
Note for l = n-1 (1s, 2p, 3d…) there is one maximum, at r = n2a
Angular functions
- orbitals
2p orbitals 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
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)
e+
e-
µ =
ve
π r 2 with
2π r
mvr = 
µ B =
e
(Bohr Magneton) 2m
So moment = mlµB
Applied B field shifts energy (dipole-field interaction) so ml states are split apart by µBB 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 sz = ±  but µ z = − 2.000232
2
Spin angular momentum S (cf L)
e
s
) 2m
z
Spin magnetic quantum number ms (cf ml)
Total combined angular momentum J
  
J = L+S
J=
(
)
j j +1 
j= l±
Combining L and S leads to correct number and splitting of emission lines (first seen by Irishman Thomas Preston)
1
2
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 ↓ ±

2
So maximum of two electrons in each state one ↑ and the other↓
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)
Z>1 The periodic table of elements
n=1 n=2 ……
…….
l=0
l=1
l=2
l=3
s state – 2 electrons
p state – 6 electrons
d state – 10 electrons
f state – 14 electrons
Gives the basic structure of the
Periodic Table of the elements
Periodic Table of Elements
1-7s
4-5f
3-6d
2-7p
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.
X-Ray spectra
X-Rays are emitted by impact of high
energy electrons on elements
λmin =
hc
eV
Continuous spectrum due Bremsstrahlung & other
scattering processes
Molybdenum spectrum shown
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
(
)
f ∝ Z −1
2
The first time Z was
spectroscopically
determined…
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)
2
3
ΔE =
4
(
)
Ryd Z − 1
Which agrees very closely with Moseley’s experiment.
Actually the most important early evidence for nuclear model of atom!