Lecture 18: Orbitals of the Hydrogen Atom The material in this lecture covers the following in Atkins. The structure and Spectra of Hydrogenic Atoms 13.2

Lecture 18: Orbitals of the Hydrogen Atom The material in this lecture covers the following in Atkins.

The structure and Spectra of Hydrogenic Atoms 13.2 Atomic orbitals and their energies (c) Shells and subshells (b) s-orbitals (c) Radial distribution function (d) p-orbitals (e) d-orbitals Hydrogenic atomic orbitals

Lecture on-line Hydrogenic atomic orbitals (PDF Format) Hydrogenic atomic orbitals (PowerPoint)Handout for this lecture

Audio-visuals on-line key hydrogen orbitals (PowerPoint)(From the Wilson Group,***) key hydrogen orbitals (PDF)(From the Wilson Group,***) Vizualization of atomic hydrogen orbitals (PowerPoint)(From the Wilson Group,***) Vizualization of atomic hydrogen orbitals (PDF)(From the Wilson Group,***) Slides from the text book (From the CD included in Atkins ,**) Interactive Hydrogen Orbital Plots (For Mac users only) ( Visualizes all the angular and radial wavefunctions of the hydrogen atom, *****)

Orbitals of Hydrogenic Atom

The orbitals for the hydrogenic atom are given byψnlm(r,φ,θ)=R(r)nlYl,m(φ,θ) ; n=1,2,3 ... l< n-1; m= -l, -l+1, ....l-1, l

Where

Yl,m(φ,θ)=2l+14π

(l−|m!|(l+|m!|)

Pl|m|(cosθ)×exp[imφ]

are eigenfunctions to L2 and Lz

With the radial part given as

Rnl(r)=Nnlρn

⎛ ⎝ ⎜

⎞ ⎠ ⎟ lLn,l(r)e

−ρ/2n

Hydrogen Levels

n =1, l=0ψn,l,m(φ, θ) =Rnl( )r Y ,l m((φ, θ)

R1,0(r) = 2Zao

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 / 2

e−ρ / 2

No nodes. R1,0 (r) everywhere positive

For l = 0 we have m = 0;

Yo,o =14π

Value of Yoo is uniform oversphere

1s−orbital

Orbitals of Hydrogenic Atom...prop.dens.

The probability densityPnlm(r,θ,φ)=ψnlm(r,θ,φ)*ψnlm(r,θ,φ)

A constant-volume electron-sensitivedetector (the small cube) gives its greatestreading at the nucleus, and a smallerreading elsewhere. The same reading isobtained anywhere on a circle of givenradius: the s orbital is sphericallysymmetrical.

P1,00(r,φ.θ) = 2Zao

⎛

⎝ ⎜

⎞

⎠ ⎟ 3/2

e−ρ/2Yoo* ×

2Zao

⎛

⎝ ⎜

⎞

⎠ ⎟ 3/2

e−ρ/2Yoo =1π

Zao

⎛

⎝ ⎜

⎞

⎠ ⎟ 3e−ρ

Representations of the1s and 2s hydrogenicatomic orbitals in termsof their electron densities(as represented by thedensity of shading).

The probability density Pnlm(r,θ,φ)=ψnlmr,θ,φ)ψnlmr,θ,φ)

P1,00(r)= 1π

Zao

⎛

⎝ ⎜

⎞

⎠ ⎟ 3e−ρ

P200(r)=1

32πZao

⎛

⎝ ⎜

⎞

⎠ ⎟ 3(2−

12

ρ)2e−ρ/2

Orbitals of Hydrogenic Atom...prop.dens.


A constant-volume electron-sensitivedetector (the small cube) gives its greatestreading at the nucleus, and a smallerreading elsewhere. The same reading isobtained anywhere on a circle of givenradius: the s orbital is sphericallysymmetrical.

P1,00(r, φ.θ) = 2Z

ao

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 / 2

e−ρ / 2Yoo ×

2Zao

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 / 2

e−ρ / 2Yoo =1π

Zao

⎛ ⎝ ⎜

⎞ ⎠ ⎟3

e−ρ

Radial probability density


The radial distribution function P gives the probability that the electronwill be found anywhere in a shell of radius r. For a 1selectron in hydrogen, P is amaximum when r is equal tothe Bohr radius a0. The valueof P is equivalent to the readingthat a detector shaped like aspherical shell would give asits radius was varied.



The probability of finding the electronbetween r and r + dr is :

Pnl(r)dr = Pnlm(r, φ, θ)dvo

π∫

o

2π∫

= Pnlm(r, φ, θ)rsinθdθdφr2dro

π∫

o

2π∫

= |Rnl((r) |2 Ylm(φ, θ) |2 sinθdθdφr2dr

o

π∫

o

2π∫

= 1 × |Rnl((r) |2 r2dr

Thus the radial probability density is

Pnl(r) =|Rnl((r) |2 r2



The radial probability density is

Pnl(r)=|Rnl((r)|2 r2

The most probable radius corrspondsto the maxima for Pnl(r).

For 1s

rmax=aoZ



Orbitals with increasing l are called

ψ 1n m =Rn1(r)Y1m(θ, φ) p- orbitals m= - 1, 0, 1

ψn2m = Rn2(r)Y2m(θ, φ) d - orbitals m = - 2, -1, 0,1,2

ψn3m = Rn3(r)Y3m(θ, φ) f - orbitals m = - 3, -2, -1,0,1, 2, 3


Rnl is a solution to

−h2

2μ{δ2Rnl( )r

δ2r+2r

δRnl( )rδr

) + {−Ze

4πεor+

h2l(l+ 1)2μmr2

}Rnl( )r =ERnl( )r

Veff

As l increases the centrigugal term h2l(l+1)

2μmr2

becomes more repulsive near nucleus at small r. Hence Rnl(r) tend to zero as r→ o with increasing spead as l becomes larger

Orbital near nucleus


Close to the nucleus,p orbitals are proportional to r,d orbitals are proportional to r2,and f orbitals are proportional to r3.Electrons are progressively excludedfrom the neighbourhood of the nucleusas l increases. An s orbital has afinite, nonzero value at the nucleus.

Orbital near nucleus


The variation of the mean radius of ahydrogenic atom with the principal andorbital angular momentum quantumnumbers. Note that the mean radius lies inthe order d < p< s fo r a give n valu e o f n.

Mean radius

The mean radius is given as the espectation value

< r >= ψnlm∫ rψnlmdv

For the hydrogenic atom

< r >nl=n2 1+12(1−

l(l+ 1)n2

⎧ ⎨ ⎩

⎫ ⎬ ⎭

ao

Z


The boundary surface of an s orbital,within which there is a 90 per centprobability of finding the electron.


ψnlm(r, φ, θ) = R(r)nlYl,m(φ, θ)

n = 2, l = 1

ψ211(r, φ, θ) = R(r)21Y1,1(φ, θ) = -R(r)213

8π ⎛ ⎝

⎞ ⎠

1/ 2

sin θeiφ

m =−1, 0, 1

ψ210(r, φ, θ) = R(r)21Y1,0(φ, θ) = R(r)213

4π ⎛ ⎝

⎞ ⎠

1/ 2

cos θ

ψ21−1(r, φ, θ) = R(r)21Y1,-1(φ, θ) = R(r)213

8π ⎛ ⎝

⎞ ⎠

1/ 2

sin θe−iφ

Let us now look at the angular part of the p - functions

Angular part


ψ210(r, φ, θ) = R(r)213

4π ⎛ ⎝

⎞ ⎠

1/ 2

cos θ l = 1,m = 0

x

y

z

L2 =h2l(l +1)

Lz=hml

m=0

The electron has an angular momentum r L

such that L∗L =h2l(l+ 1)The z- component of

s L is Lz =o

Angular part

ψ211(r, φ, θ) = -R(r)213

8π ⎛ ⎝

⎞ ⎠

1/ 2

sin θeiφ


x

y

z

L2 =h2l(l +1)

Lz=hml

x

yz

r v

v L =v

r ×v v

l = 1,m = 1



s L is Lz =h

Angular part


ψ21−1(r, φ, θ) = R(r)213

8π ⎛ ⎝

⎞ ⎠

1/ 2

sin θe−iφ l = 1,m = -1

x

yz

L2 =h2l(l +1)

Lz=hml

m=-1



s L is Lz =−h

Angular part

Plotting angular part of

ψ210 ( ,r φ, θ) = ( )R r 2134π ⎛ ⎝

⎞ ⎠

1/ 2cos θ =2pz


R =k cos θ

Θ +

-

Draw vector r R of length

kcosθ through each point(θ, φ) on sphere

Draw surface through all vectors R

Real orbitals

ψ211(r, φ, θ) = -R(r)213

8π ⎛ ⎝

⎞ ⎠

1/ 2

sin θ[cos φ+i sin φ]


ψ21−1(r, φ, θ) = R(r)213

8π ⎛ ⎝

⎞ ⎠

1/ 2

sin θ[cos φ−i sin φ]

The remaining two orbitals are complex

However by linear combinations we can get the real orbitals

2py =i2{ψ211 +ψ21−1} = ( )R r 21

38π ⎛ ⎝

⎞ ⎠

1/ 2

sinθ sinφ

2px =12{ψ211 - ψ21−1} = ( )R r 21

38π ⎛ ⎝

⎞ ⎠

1/ 2

sinθ cos φ

Angular part

2py =i2{ψ211 +ψ21−1} = ( )R r 21

38π ⎛ ⎝

⎞ ⎠

1/ 2

sinθ sinφ

2px =12{ψ211 - ψ21−1} = ( )R r 21

38π ⎛ ⎝

⎞ ⎠

1/ 2

sinθ cos φ


The orbitals 2px and 2py have the same energies and eigenvalues to

L2 as ψ211 andψ21-1 . However only the latter are eigenfunctions of Lz

+

-x

z

+-

y

2py 2px

Angular part


The boundary surfacesof p orbitals. A nodalplane passes through the nucleus andseparates the two lobesof each orbital.The dark and light areas denoteregions of opposite signof the wavefunction.

Angular part


A Grotrian diagram that summarizes theappearance and analysis of the spectrum ofatomic hydrogen. The thicker the line, themore intense the transition.

In transitions between different energy levels ofthe hydrogenic atom the following selectionrules apply

Δl=±1 Δm= o, ±1

Transitions

What you need to learn fromthis lecture about the hydrogen atomfor the quizz and the final exam

The definition of the radial distribution function

P(r)=r2R(r)2 and how it is used to calculate the

expaction values <rn >

The behaviour of Rnl(r) nearthe nucleus (Fig 13.16)

Understand how the balance between kinetic energy and potential energy shapes RnlFig.13.10

Selection rules for electronictransitions in thehydrogen atom

Δl=±1 Δm= o, ±1

Memorize the relation between real p-orbitals (px,py,pz) and imaginary p-orbitals (po,p−1,p1).Understand the physical difference. Understandplots of (px,py,pz).

Same for d and f great,but not required

Appendix:Orbitals of Hydrogenic Atom..why nodes

R20 (r) = 1

2 2Zao

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 / 2

(2 −12ρ)e−ρ / 4

n =no, l=0ψn,l,m(φ, θ) =Rnl( )r Y ,l m((φ, θ)

R30 (r) = 1

9 3Zao

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 / 2

(6 −2ρ+19ρ2 )e−ρ / 6

One node at

(2−12ρ) =0

with ρ =2 /Zr ao; node : 2ao / Z

two nodes at

(6−2ρ +19ρ2 ) =0

with ρ =2 /Zr ao; nodes : 1.9ao /Zand 7.1 /Zr ao

For l = 0 we have m = 0;

Yo,o =14π

n =no, l=0ψno ,o,o (φ, θ) =Rno

,o ( )r Y ,o o((φ, θ)

Thus any two s - orbitals Rn'oYoo and RnoYoo

must be orthorgonal

Rn'oYoo0

∞∫

0

π∫

0

2π∫ Rn'oYoodv =0

Why has Rnl(r) n-1 nodes ?

We have seen previously that any two independent solutions to the Schrödinger equation must be orthogonal :

ψ i∫ ψjdv = ∂ij


Rn'oYoo0

∞∫

0

π∫

0

2π∫ Rn'oYoodv =0

Yooo

π∫

o

2π∫ Yoo sinθdθ dφ Rn'o

o

r

∫ (r)Rno(r)r2dr =0

1 Rn'oo

r

∫ (r)Rno(r)r2dr =0

Yoo normalized

The R1o function is positiveevery where

R1,0(r) = 2Zao

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 / 2

e−ρ / 2


Thus for R2o to be orthogonalto R1o

R1oo

r

∫ (r)R2o(r)r2dr =0

R20 must have positive and negative regions

R20 (r) = 1

2 2Zao

⎛ ⎝ ⎜

⎞ ⎠ ⎟3 / 2

(2 −12ρ)e−ρ / 4

For R3o to be orthogonal toR1o and R20 two nodes arerequired etc...


e

1

e

2

e

3

e

4

e

5

Appendix: Drawing Orbitals of Hydrogenic AtomAn asideHow do we plot Ylm ?

Consider a unit sphere withradius 1. Draw a large number ofunit vectors from the origin ofthe sphere in different directions(e.i. different θ, φ)

r e1 ,

r e2 ,

r e3 , r

e4 ... r en

Calculate the value of Ylm(φ, θ)at the position of each

r en

( . e i for eachφn, θn). Construct

the vectors: R1 , R2 , R3 , .. Rn

where: r Rn =

r enYlm(φ, θ)

Draw Rn

R

2

R

3

1

R

R

4

R

5

An aside

R

2

R

3

1

R

R

4

R

5

Draw a surface through the

endpoints of all r R n .

This surface represents Ylm(θ, φ)

Appendix: Drawing Orbitals of Hydrogenic Atom

An aside

For Yoo(φ, θ) we have

r Rn =

12 π

r en for all n

We have that Yoo(φ, θ) is spherical. . e u the same for allφn, θn

THus Yoo representing the angularpart of a ns function is representedby a sphere

ns - orbital

Appendix: Drawing Orbitals of Hydrogenic Atom

Documents

Lecture 18: Orbitals of the Hydrogen Atom The material in this lecture covers the following in Atkins. The structure and Spectra of Hydrogenic Atoms 13.2