ms310 quantum physical chemistry

Ch 9. The Hydrogen AtomCh 9. The Hydrogen Atom

MS310 Quantum Physical Chemistry

- Historical hydrogen atom model (Plum pudding model, - Historical hydrogen atom model (Plum pudding model, Rutherford model) vs. quantum mechanical modelRutherford model) vs. quantum mechanical model

- Formulate the Schrödinger equation for hydrogen atom - Formulate the Schrödinger equation for hydrogen atom and solve itand solve it

- Study the energy level and orbitals of hydrogen atomStudy the energy level and orbitals of hydrogen atom


9.1 Formulating the Schrödinger equation9.1 Formulating the Schrödinger equationRutherford vs shell model electrons are confined in spherical shells centered at nucleus orbit the nucleus, accelerating motion and energy radiation → atom is ‘not’ stable!This problem is solved by Quantum approach.We consider the Coulomb potential

re

re

rV0

2

0

2

4||4)(

H atom : 1 proton + 1e–

H-like atom : Z proton + 1e– ex) He+

Hamiltonian of hydrogen atom is

re

mm NN

ee

KeK

0

22

22

2

nucleus,,

422

VEEH

x

z

y

e–

e+


Focus on the internal motion(center of mass motion : translation)Therefore, Schrödinger equation is written as

),,(),,(4

]),,(sin1)),,((sin

sin1)),,((1[

2

0

2

2

2

222

2

2

rErr

e

rr

rrr

rr

rrme

(in generally, we must use μ instead of me. However, in the hydrogen atom case, both are almost same and this book use me.)

Use the center of mass(already discuss in chapter 7), we can divide it by 2 equations.

motioninternal :2

motionC.O.M. :2

)(22

22

COMCOMCOM2

COM

2

COMtotalCOMtotal

22

2COM

2

rrrrr

rr

r

EV

EM

EEE

rVM

H


9.2 Solving the Schrödinger equation for the Hydrogen 9.2 Solving the Schrödinger equation for the Hydrogen atomatom

Use the separation of variable : ψ(r,θ,φ) = R(r)Θ(θ)Φ(φ)We know the form of from chapter 72l

)φ,θ(E)φ,θ(YlI2

1]

φ)φ,θ(Y

θsin1

)θ

)φ,θ(Yθ(sin

θθsin1

[rμ2 Y

22

2

220

2

==+ ∂∂

∂∂

∂∂

Rewrite the Schrödinger equation using the angular momentum

)()()(

)(4

)()()()(ˆ)(2

1))(()()(2 0

22

22

2

2

rER

rRr

elrR

rmdrrdR

rdrd

rm ee

Focus on the radial part(we already know the angular solution)

)()(]42

)1([))((2 0

2

2

22

2

2

rERrRr

erm

lldr

rdRr

drd

rm ee


coefficient of second term : effective potential

First term : centripetal potential, related to 1/r2

Second term : coulomb potential, related to -1/r

Unless the l=0 case, centripetal potential is dominant → electron of p, d, f orbital(l>0) far from nucleus than electron of s orbital(l=0)

re

rmll

rVe

eff0

2

2

2

42)1()(

9.3 Eigenvalues and eigenfunctions for the total energy9.3 Eigenvalues and eigenfunctions for the total energy


We can divide the radial part equation i) V=0

)()( where 0)()1(12

introducetoconvenient

0)(2)1(2

,22

2

2

,

2

22

2

rRRRll

drd

dd

kr

k

rREm

rll

drd

rdrd

lEll

lEe

cos3sin13)(

cossin)(

sin)(

232

21

0

j

j

j

Solution : spherical Bessel function

2/1

0

22/1

2

,22,

2

,,

,,0

2

22

2

2

24 and8

econveniencforquantitiesessdimensionlintroduceAgain

0)()(2

)()( introducetoconvenient

)()(4

)1(12

Eme

rEm

rurVEm

drud

rrRru

rERrRr

erll

drd

rdrd

rm

ee

lEelE

lElE

lElEe

eff

) inl(polynomia) inlexponentia(decaying)()( 2/

,

feu lE

)(/)( ,2/

,,, ln

llnlnln LeNrurR

ii) V≠0

The equation becomes to

Equation Lagurre Associated:0)(]41)1([ '

,'''

2

22

lEull

dd

Again introduce dimensionless quantities for convenience


Energy of hydrogen atom :

Define the constant

: 0.529Å for hydrogen atom : Bohr radiusTherefore, energy is

...4,3,2,1,8 222

0

4

nnh

emE e

n

2

20

0 emh

ae

...4,3,2,1,eV 60.13J10179.28 22

18

220

20

2

nnnna

eEn

For n>5 state, states are in the narrow range, 0 to -1x10-19 J.Potential of H atom : very narrow for first few states, but very wide for large nAs known from a particle in a box : energy spacing is inverse of the square of box length


Energy : depends on only the principle quantum number nHowever, wavefunction depends on the 3 quantum numbers, n, l, and ml

The relationship is given by n : 1, 2, 3, 4, … l : 0, 1, 2, 3, …, n-1 ml : 0, ±1, ±2, …, ±l (existence of these quantum numbers are from boundary condition)

Radial function R(r) : product of exponential function with a polynomial, dimensionless variable r/a0

Functional form of radial function : depends on the quantum numbers n and l.


First few radial functions Rnl(r) are given by

0

0

0

0

0

0

320

223

032

320

2

0

23

031

320

2

0

23

030

2

0

23

021

2

0

23

020

23

010

)1(3081

4)(:2,3

)6()1(69

1)(:1,3

)21827()1(381

2)(:0,3

)1(241)(:1,2

)2()1(8

1)(:0,2

)1(2)(:0,1

ar

ar

ar

ar

ar

ar

ear

arRln

ear

ar

arRln

ear

ar

arRln

ear

arRln

ear

arRln

ea

rRln


Therefore, wavefunction of hydrogen atom, ψnlml is given by

iar

l

ar

l

ar

l

ar

l

eear

armln

ear

armln

ear

armln

ea

rmln

sin)1(8

1),,(:1,1,2

cos)1(241),,(:0,1,2

)2()1(241)(:0,0,2

)1(1)(:0,0,1

0

0

0

0

2

0

23

0121

2

0

23

0210

2

0

23

0200

23

0100


iar

l

iar

l

ar

l

iar

l

ar

l

ar

l

eear

armln

eear

armln

ear

armln

eear

ar

armln

ear

ar

armln

ear

ar

armln

22320

223

0232

320

223

0132

2320

223

0320

320

2

0

23

0131

320

2

0

23

0

21

310

320

2

0

23

0300

sin)1(162

1),,(:2,2,3

cossin)1(81

1),,(:1,2,3

)1cos3()1(681

1),,(:0,2,3

sin)6()1(81

1),,(:1,1,3

cos)6()1()2(811),,(:0,1,3

)21827()1(381

1)(:0,0,3

0

0

0

0

0

0


Hydrogen wave function


These functions are called to both eigenfunction and H atom ‘orbitals’

There are some property of orbital 1) Letter s, p, d, f are used to denote l = 0, 1, 2, 3 2) ψ100(r) : 1s orbital or wave function 3) all 3 wave functions with n=2, l=1 : 2p orbitals 4) wavefunction is real when ml=0, complex when otherwise

Wavefunction is normalized to generate the probability density.(postulate 3)

Energy of H atom : degenerated

- n=1 : no degeneracy - n=2 : 4-fold degeneracy - n=3 : 9-fold degeneracy

21

0

1

0)12()state each( nl

n

l

n

l

l

lml


Use the superposition principle : if y1 and y2 are solutions of DE, then c1y1+c2y2 is also solution of DE. → can make the complex functions to real

cos)6()1(81

2),,(

sinsin)6()1(81

2),,(

cossin)6()1(81

2),,(

cos)1(241),,(

sinsin)1(241),,(

cossin)1(241),,(

0

0

0

0

0

0

320

2

0

23

03

320

2

0

23

03

320

2

0

23

03

2

0

23

02

2

0

23

02

2

0

23

02

ar

p

ar

p

ar

p

ar

p

ar

p

ar

p

ear

ar

ar

ear

ar

ar

ear

ar

ar

ear

ar

ear

ar

ear

ar

z

y

x

z

y

x


2sinsin)1(281

1),,(

2cossin)1(281

1),,(

sincossin)1(81

2),,(

coscossin)1(81

2),,(

)1cos3()1(681

1),,(

2320

223

03

2320

223

03

320

223

03

320

223

03

2320

223

03

0

0

22

0

0

0

2

ar

d

ar

d

ar

d

ar

d

ar

d

ear

ar

ear

ar

ear

ar

ear

ar

ear

ar

xy

yx

yz

xz

z

Real form : more convenient to visualize the chemical bondHowever, real form is not an eigenfunction of

zl


Experimental emission spectra is given by

Use the reduced mass instead of me, frequency of spectral line is given by

Sometimes, we use wave number instead of frequency

Rydberg constant : mee4/8ε02h3c : 2.180x10-18 J , 109737 cm-1

Reduced mass of H : 0.05% greater than me

Spectral line of H is given by

)(1finalinitial EE

h

)11(8 2232

0

4

finalinitial nnhe

1~ c

9.4 The hydrogen atom orbital9.4 The hydrogen atom orbital


Bohr model : electron orbit around the nucleus and only certain orbits allowed

Probability : proportional to ψ*(r,θ,φ)ψ(r,θ,φ)dτ

Our focus is on the 1) wave function ψnlml(r,θ,φ) 2) probability of finding electron ψ2

nlml(r,θ,φ)sinθdrdθdφ 3) define radial distribution function and

Ground state of H atom : ψ100(r)

Plot it : need 4 coordinate(x, y, z and P(x,y,z))

0

23

0100 )1(1)( a

r

ea

r


a) 3D plot on x-y half plane b) contour plot on x-y half plane red : high probability, blue : low probability


Plot of 1s, 2s, and 3s orbital


Plot R(r) vs r : 1s, 2s, 2p, 3s, 3p, and 3d orbital


Ground state : no radial nodesQuantum number increase : # of nodes increaseWhen 2s and 3s, node has constant r : spherical ‘nodal surface’Then, what about the node of another orbitals? → l>0 : not spherical symmetry : ‘angular shape’ of orbital

See contour of 2py, 3py, 3dxy and 3dz2 orbital → we can see the angular nodal surfaceNodal surface of 2py : y=0m no radial nodes

Generally, l nodal surfaces in angular part and n-l-1 radial nodal surfaces, n-1 total nodal surfaces

3py : additional nodal plane x=0 : radial node3dxy : 2 nodal planes intersect the z axis3dz2 : 2 ‘conical’ nodal surfaces, rotating the z axis


Ex)9.3

Locate the nodal surfaces in

Sol) consider the radial and angular part separately.

Angular part : cos θNode : cos θ = 0, θ = π/2It means the plane z=0 in Cartesian coordinate

Radial part :

Node depends on only (exponential term cannot be zero) → r=0 and r=6a0

r=0 : a point → no meaning, r=6a0 : a surfaceTherefore, there are 1 angular and 1 radial node.It is same as the general result(l angular nodes and n-l-1 radial nodes)

cos)6()1()2(811),,( 03

20

2

0

23

0

21

310ar

ear

ar

ar

20

2

0

6ar

ar

0320

2

0

)6( ar

ear

ar


9.5 The radial probability distribution function9.5 The radial probability distribution function

See the ψ2n00(r,θ,φ) : n=1,2 and 3

Maxima is at r = 0

Consider the ψ2nlml(r,θ,φ) in general

case(l>0) → centripetal barrier, nonzero angular momentum → wavefunction is not spherically symmetric. : p and d orbitals


Ex)9.4

a) Where is the maximum probability point? b) Assume nucleus diameter of H is 2x10-15m. Then, probability of electron of 2s orbital is in the nucleus?

Sol)

a) The point : maximum value of ψ*(τ)ψ(τ)dτ

Only see the and differentiate it

However, r cannot be negative → consider the ρ=0

Therefore, maximum point is ρ=0 → r=0

02

0

23

0200 )2()1(

321)( a

r

ear

ar

02

0

3

0200

*200 )2()1(

321)()()( a

r

ear

arrrP

0

2 ,)2()(ar

ef

4,2,0)86()( 2 e

df

44)4(,0)2(,4)0( efff


b) Result of a) : unphysicalProbability is given by

Assume the integrand is constant on the interval(rnucleus : small)

Approximately, because of small rnucleus Finally we can obtain

Therefore, probability of finding the electron in the nucleus is essentially zero.

nucleusr

ar

drear

rdda

P0

2

0

2

0

2

0

3

0

0)2(sin)1(32

1

153

0

32

0

23

0

100.9)(61

34])2[(4)1(

321

0

a

rre

ar

aP nucleus

nucleusa

rnucleus

nucleus

3nucleus

ar

2

0

2nucleus3

0

r

0

2ar

2

0

2nucleus3

0

r

0

ar

2

0

2

0

2

0

3

0

r34]e)

ar2[(4)

a1(

321

drr]e)ar2[(4)

a1(

321dre)

ar2(rdsind)

a1(

321P

0

nucleus

nucleus

0

nucleusnucleus

0

1,22 0

0

2

ar

nucleusnucleus

ea

r


Plot of a03R2(r) vs r/a0


R(r) cannot describe the ‘real’ distribution! → radial probability depends on the ‘summation over all θ and φ’

For 1s orbital,

Generally, introduce the ‘radial distribution function’ P(r)

We can determine the most probable position of electron.

Understand the difference between radial distribution and probability density function is very important.

0

22

0

),,(sin)( drrdddrrPlnlm

drera

drerdda

drrP ar

ar

00

22

300

22

2

030

4sin1)(

drrRrdrrP 22 )]([)(

9.6 The validity of the shell model of an atom9.6 The validity of the shell model of an atom


See the plot of radial distribution function


Classical shell model

Quantum distribution of 1s, 2s, and 3s

Classical shell model is not valid any more.However, we can see the most dense point of 1s orbital is very less intensity when 2s and 3s orbital! → classical shell model is useful although reducing a complex function to a single number is unwise.


- Result of solving the Schrödinger equation for hydrogen atom is exactly equal to experimental data.

- Shape of orbital is changed by the quantum number and probability of finding electron depends on the shape of orbital.

- There are n-l-1 radial nodal surfaces and l angular nodal surfaces, # of total nodes is n-1 for nth state.

Summary Summary

ms310 quantum physical chemistry

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