# And L

advertisement ## The Hydrogen atom

-

Solution of Schroedinger equation

-

3 spatial quantum numbers

-

Probability and ‘Orbitals’

-

Selection rules for transitions

-

Zeeman Effects, electron spin

-

4 quantum numbers

-

Combining angular momenta

## Hydrogen Atom Re-visited

First real test of Schroedinger Equation:

Recall Bohr model:

E n

-1

/n

2

, via one quantum number

n

(also, quantised angular momentum L via same n)

L

Successes:

=

Hydrogen line emission spectra

also… size & stability of atoms, Moseley’s Law for Z, etc. n

Problems: Full 3D & spherical symmetry?

Magnetic effects in spectra

Z > 1 atoms’ spectra

Comparison with Bohr model: same expression for

E n

via “principal” quantum number

n

Schroedinger Equation for H atom

Express wavefunction as product

(c.f 2-D box)

Ψ

(

r

,

θ

,

φ

)

=

( )

Θ

( )

Φ

Leads to 3 separate equations

2

2

mr

2

d dr

r

2

dr

+

2

2

mr

2

+

( )

1 sin

θ

d d

θ

⎝ sin

θ

d

Θ

d

θ

+

( )

m l

2 sin

2

θ

( )

=

=

0

d

2

Φ

d

φ

2

+

m l

2

Φ =

0

R solutions are

e

α r

( × r polynomial)

type, Θ are functions of cos & sin θ

Φ solutions are

exp(im l

ϕ

)

type, where

m l

is an integer

Solutions specified by three quantum numbers:

n, l, m l

Schroedinger Equation for H atom (cont.)

Angular momentum - is quantised

Solution shows

L

=

(c.f. angular momentum in lecture 11)

( )

Also Φ must be single valued and

Find

L z

=

m l

(

m l

=

Φ

0,

±

1,

±

= Φ

(

φ +

2

π

)

2...

±

l

)

Note that

L

can be zero

(unlike Bohr model)

And

L z

< L

(unless L=0)

Exists an Uncertainty relation between

L x

L y

L z

Bohr – plane, but must have Δ p normal to plane…

Also seen via

But

⎡⎣

L

2

,

L z

⎤⎦

L x

L y

L y

L x

=

⎡⎣

L x

,

L y

⎤⎦

=

i

L z

(try it! – commutator)

=

so no uncertainty between L and L z

& can know both

‘Semi-classical’ vector

pictures for l=2

Schroedinger Equation for H atom (cont.)

Solutions specified by three quantum numbers:

n, l, m l

Energy Levels are

E n

and the same as the Bohr model!

with

n = 1, 2, 3, 4, ….

Solutions show

l = 0, 1, 2, 3… n-1

and |

m l

| ≤

l

Notation:

s, p, d, f…

for

l = 0, 1, 2, 3, ….

So, possible states of electron:

(

c.f. 3-D box, Lect. 9)

E n

= −

me

4

8

ε

2

o h

2

Note: E n

i.e. l, m

does not depend on

l, m l l states are degenerate –

but holds only for H atom

Electron states in hydrogen atom - orbitals

( )

dr

=

ψ

2

4

π

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 = n

2 a

Angular functions

- orbitals

The (m

l

) sub-shells combine to give

spherically symmetric (

l

) shells - Spherical

‘harmonics’

(complete shells…) e.g. for 2p-states, cos θ & sin θ dependence

Squared (probability) & add - gives 1

Transition of electrons between orbitals

Selection rules

for transitions:

Δ

l

= ± 1

And

Δ m l

= 0 or ± 1

All others are ‘forbidden’

Conservation overall of angular momentum, photon has angular momentum of

l

determines the magnitude of

L

, m

l

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

2

π

r

π

r

2 with

mvr

=

µ

B

=

e

2

m

(Bohr Magneton)

So moment =

m l

µ

B

Applied B field shifts energy

(dipole-field interaction)

so

m l

states are split apart by µ

B

B

### Zeeman Effect

In non-zero magnetic field B each

l

-level splits into 2

l

+1

m

l

sub-levels of slightly

different energy, depending

on B

Spectra show three lines

instead of one; only three

because of further selection

rule:

Δ

m l

= ± 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 11 !

Electron has spin angular momentum

s z

= ±

but

2

Spin angular momentum

S (cf L)

Spin magnetic quantum number

m s

(cf m l

)

µ

z

= −

(

2.000232

)

e

2

m s z

Total combined angular momentum

J

J

= +

S

J

=

( )

j

=

l

±

1

2

Combining

L

and

S

leads to correct number and splitting of emission lines

(first seen by Irishman Thomas Preston)