lecture 18: orbitals of the hydrogen atom the material in this lecture covers the following in...
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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.
Orbitals of Hydrogenic Atom
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
Orbitals of Hydrogenic Atom
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.
Radial probability density
Orbitals of Hydrogenic Atom
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
Radial probability density
Orbitals of Hydrogenic Atom
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
Radial probability density
Orbitals of Hydrogenic Atom
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
Orbitals of Hydrogenic Atom
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
Orbitals of Hydrogenic Atom
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
Orbitals of Hydrogenic Atom
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
Orbitals of Hydrogenic Atom
The boundary surface of an s orbital,within which there is a 90 per centprobability of finding the electron.
Orbitals of Hydrogenic Atom
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
Orbitals of Hydrogenic Atom
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
Orbitals of Hydrogenic Atom
x
y
z
L2 =h2l(l +1)
Lz=hml
x
yz
r v
v L =v
r ×v v
l = 1,m = 1
The electron has an angular momentum r L
such that L∗L =h2l(l+ 1)The z- component of
s L is Lz =h
Angular part
Orbitals of Hydrogenic Atom
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
The electron has an angular momentum r L
such that L∗L =h2l(l+ 1)The z- component of
s L is Lz =−h
Angular part
Plotting angular part of
ψ210 ( ,r φ, θ) = ( )R r 2134π ⎛ ⎝
⎞ ⎠
1/ 2cos θ =2pz
Orbitals of Hydrogenic Atom
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]
Orbitals of Hydrogenic Atom
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
Orbitals of Hydrogenic Atom
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
Orbitals of Hydrogenic Atom
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
Orbitals of Hydrogenic Atom
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
Appendix:Orbitals of Hydrogenic Atom..why nodes
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
Appendix:Orbitals of Hydrogenic Atom..why nodes
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...
Appendix:Orbitals of Hydrogenic Atom..why nodes
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