Last Lecture• Observed spectra and Bohr’s energy levels

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E =p2

2m−kZe2

r⇒ E = −

12k 2Z 2me4

n2h2= −13.6eV Z 2

n2

For Rydberg atoms the effect of the screening of other electrons can be accounted for by substituting (Z-1)2 to Z2

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1λ

= Z 2R 1n f2 −

1ni2

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Ei − E f =hcλ

=k 2me4

2h2Z 2 1

n f2 −

1ni2

⇒

1λ

=k 2me4

4πh3cZ 2 1

n f2 −

1ni2

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Hydrogen atom wavefunction• We have to extend the wavefunction concept from 1D to 3D

• Simplification: Hydrogen has spherical symmetry

• Ψ(x, y, z) is converted to Ψ(r,θ,ϕ)

• We need three quantum numbers to represent 3D states – Radial distance from nucleus– Azimuthal angle around nucleus– Polar angle around nucleus

• Quantum numbers are integers (n, l, ml), and the spin quantum number completes the picture

• We can’t say exactly where the particle is, but we can tell you how likely is to find the particle in a particular location.

The Schroedinger equation for H atom

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Z=1

Spherical coordinates

Laplace operator or Nabla square in spherical coordinates

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−h2

2µ∇2 + (U − E)

Ψ = 0

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∇2 =1r2

∂∂r

r2 ∂∂r

+

1r2 sinθ

∂∂θ

sinθ ∂∂θ

+

1r2 sin2θ

∂ 2

∂ϕ 2

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∇2 =∇ ⋅∇

Separating variables

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Z=1

radial equation colatitude equation

=-m2

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Ylm (θ,ϕ) = Plm (θ)Fm (ϕ)radial equationspherical harmonics

azimuthal equation (depends only on φ)

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sin2θR

ddr

r2 dRdr

+sinθP

ddθ

sinθ dPdθ

+2µr2 sin2θ

h2E −U(r)( ) = −

1Fd2Fdφ 2

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1Rddr

r2 dRdr

+2µr2

h2E −U(r)( ) = −

m2

sin2θ−

1P sinθ

ddθ

sinθ dPdθ

= l(l +1)

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Rnl (r)

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Ψ = R(r)P(θ)F(φ)

The solutions and quantum numbers

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Bound state energy spectrum is quantized,

En = −E0

n2 , n =1,2,3,K

Principal quantum number

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Angular momentum is quantized (depends on n),

L = l(l +1) h2π

, l = 0,1,2,K,n −1( )

Solutions exist when these conditions are satisfied on quantum numbers

Orbital quantum number

Magnetic quantum number

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The direction of angular momentum is also quantized,

Lz = mlh

2π, ml = −l,−l +1,K,−1,0,1,K,l −1,l( )

The z component of L is quantized2l+1states

Another quantum number is needed: the spin!

Quantization of angular momentum

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p

pr

pt

r

α

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L = r ×p⇒ L = rpsinα = rpt

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E =p2

2µ+U(r) =

pr2 + pt

2

2µ+U(r) =

pr2

2µ+

L2

2µr2+U(r) =

pr2

2µ+Ueff (r)

The angular part is:

=-m2

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1Rddr

r2 dRdr

+2µr2

h2E −U(r)( ) = −

m2

sin2θ−

1P sinθ

ddθ

sinθ dPdθ

= l(l +1)

From

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−1

P sinθddθ

sinθ dPdθ

+

1F sin2θ

d2Fdθ 2

= l(l +1)

Defining the operator for L2

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(L2)op = −h21sinθ

ddθ

sinθ ∂∂θ

+

1sin2θ

∂ 2

∂θ 2

This is the angular part of the S. eq.: for a potental U(r) the angular momentum is quantized

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L = l(l +1)h

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l = 0,1,2,...,n −1

O

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(L2)opYlm (θ,φ) = h2l(l +1)Ylm (θ,φ)

Solutions• In order to have well behaved solutions only certain values of the energy

are possible (identical to Bohr’s energy levels!)• Angular wave function

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Plm (θ) = Plm (cosθ)

Fm (ϕ) = Aexp(imφ)

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l = 0,1,2,...,n −1m = 0,±1,...,±l

Associated Legendre Polynomials

Radial solution

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• |Ψ|2 = square magnitude of the wavefunction

• P(x)dx = |ψ|2dx = probability to find the particle in a infinitesimal interval dx around x

• It is normalized to 1 because we are sure that the particle must be somewhere along x

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Probability densities

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|ψ |2 dx−∞

+∞

∫ =1

The ground state probability density

• Ground state: lowest energy

• Surface of const probability is surface of a sphere.• Probability decreases exponentially with radius.

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n =1, l = 0, ml = 0

In Bohr’s model the electron is in an orbit of radius a0.The electron can be found at any distance, and the most probable is r=a0

P(r)dr is a prob. per unit volume

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P(r)dr∝ Ψ24πr2dr∝ r2e−2r / a0dr

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Hydrogen atom probabilities

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n = 2, l =1, ml = 0

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n = 2, l =1, ml = ±1

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n = 2, l = 0, ml = 0

2s-state

2p-state 2p-state n=2 Same energy, but different probabilities

http://webphysics.davidson.edu/faculty/dmb/hydrogen/

Energy is lower for 1s than 2s and from radial probability we see that 2s electrons can be on average much farther from the nucleus than 1s electrons

Why do 1s shells fill first before 2s?

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n = 3, l =1, ml = 0

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n = 3, l =1, ml = ±1

3s-state3p-state

3p-state

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n = 3, l = 0, ml = 0

n=3: two s-states, six p-states

3s-state3p-state

3p-state

…ten d-states

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n = 3, l = 2, ml = 03d-state

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n = 3, l = 2, ml = ±1

3d-state

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n = 3, l = 2, ml = ±2

3d-state

• allowed transitions follow the selection rules related to conservation of angular momentum (and also the photon has an intrinsic angular momentum)

Energy levels

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Δml = 0,±1Δl = ±1