slide 1 chapter 6 the hydrogen atom. slide 2 outline the hydrogen atom schrödinger equation the...

Chapter 6

The Hydrogen Atom

Outline

• The Hydrogen Atom Schrödinger Equation

• The Radial Equation Solutions (Wavefunctions and Energies)

• Atomic Units

• The Hydrogen Atom Wavefunctions (Complex and Real)

• Use of the Wavefunctions (Calculating Averages)

• The Radial Distribution Function

The Hydrogen Atom Schrödinger Equation

The Potential Energy

For two charges, q1 and q2, separated by a distance, r:

Force: 1 22

04

q qf

r Potential Energy: 1 2

0

( )4

q qV r

r

2

0

( )4

ZeV r

r

If the charges are of opposite sign, the potential energy, V(r),is negative; i.e. attractive.

“Hydrogenlike” Atoms (H, He+, Li2+, etc.)

Ze

-e

r

The Schrödinger Equation

2 2 22

2 2 2 2 20

1 1 1s in ( )

2 s in ( ) s in ( ) 4

Z er E

m r r r r r r

22

2V E

m

2 22

02 4

ZeE

m r

In this equation, m represents the mass of an electron (9.11x10-31 kg).Strictly speaking, one should use the reduced mass, , of an electron andproton. However, because the proton is 1830 times heavier, = 0.9995m. Therefore, many texts (including ours) just use the electron mass.

Because V = V(r), one can solve the equation exactly if the Laplacian iswritten in spherical polar coordinates, giving:

2 2 2

2 22 2 2 2

0

1 1 1 1s in ( )

2 2 s in ( ) s in ( ) 4

Z er E

m r r r m r r

2 2 2

2 22 2 2 2

0

1 1 1 1s in ( )

2 2 s in ( ) s in ( ) 4

Z er E

m r r r m r r

L2 Operator^

2 2 22

2 20

1

2 2 4

L Z er E

m r r r m r r

^

The sole dependence of this equation on or is embodied in

the L2 operator.^

We learned in Chapter 4 (The Rigid Rotor) that the Spherical

Harmonics, Yl,m(, ), are eigenfunctions of L2 .^

2 2( , ) ( 1 ) ( , )m mL Y Y ^

ˆ ( , ) ( , )z m mL Y m Y They are also eigenfunctions of Lz:^

2 2 22

2 20

1

2 2 4

L Z er E

m r r r m r r

^

One can remove the dependence of this equation on and ,

embodied in L2, by assuming that:^

( , , ) ( ) ( , )mr R r Y

^This gives:

2 2 22

2 20

1( ) ( , ) ( ) ( , ) ( ) ( , )

2 2 4m m m

L Z er R r Y R r Y E R r Y

m r r r m r r

2 2 22

2 20

1 ( 1)( ) ( , ) ( ) ( , ) ( ) ( , )

2 2 4m m m

Z er R r Y R r Y ER r Y

m r r r m r r

Because 2 2( , ) ( 1 ) ( , )m mL Y Y ^

2 2 22

2 20

1 ( 1)( ) ( , ) ( ) ( , ) ( ) ( , )

2 2 4m m m

Z er R r Y R r Y ER r Y

m r r r m r r

We can now remove the dependence on and completely.

2 2 22

2 20

1 ( 1)( ) ( )

2 2 4

Z er R r ER r

m r r r m r r

We now have the “Radial Equation” for the hydrogen atom.

The solution to this equation is non-trivial to say the least.We will just present the solutions below.

This equation must be solved subject to the boundary condition:R(r) 0 as r .

Note: In retrospect, it should not be surprising that the angular solutions of the hydrogen atom are the same as the Rigid Rotor (Chapter 4).

Neither potential energy function (V=0 for the Rig. Rot.) dependson or , and they must satisfy the same angular BoundaryConditions.

Outline



• Atomic Units




2 2 22

2 20

1 ( 1)( ) ( )

2 2 4

Z er R r ER r

m r r r m r r

When the equation is solved and the boundary condition,R(r) 0 as r , is applied, one gets a new quantum number, n,with the restriction that:

Together with the two quantum numbers that came from solutionto the angular equations, one has three quantum numberswith the allowed values:

1 , 2 , 3 , 4 ,n

0 , 1 , 2 , 1n

0 , 1 , 2 ,m

The Third Quantum Number

or, equivalentlyn n

The Radial Equation Solutions

The Radial Wavefunctions

The functions which are solutions of the Radial Equation are dependentupon both n and l and are of the form:

0

0

( )Zr

nan

ZrR r Poly e

na

where

20

0 2

4a

me

= 0.529 Å

Bohr Radius

Several of the Radial functions are:

01 0 1 0( )

Z r

aR r N e

0220 20

0

( ) 12

Zr

aZrR r N e

a

0221 21

0

( )2

Zr

aZrR r N e

a

0

2

330 30

0 0

( ) 27 18 23 3

Zr

aZr ZrR r N e

a a

0

2

331 31

0 0

( ) 63 3

Zr

aZr ZrR r N e

a a

0

2

332 32

0

( )3

Zr

aZrR r N e

a

The Energies

The energy eigenvalues are dependent upon n only and are given by:

2 4

2 2 20

1

2(4 )n

m Z eE

n

2 2

20 0

1

2 (4 )

e Z

a n

This expression for the energy levels of “hydrogenlike” atoms isidentical to the Bohr Theory expression.

However, the picture of electron motionfurnished by Quantum Mechanics is completely different from that of the semi-classical Bohr model of the atom.

En

erg

y

2

2

12 6 2 5 k J /m o l

2

Z

n

1 -k

2 -k/4

3 -k/94 -k/16

0

Show that R10(r) is an eigenfunction of the Radial equation

and that the eigenvalue is given by E1 (below).

2 2 22

2 20

1 ( 1)( ) ( )

2 2 4

Z er R r ER r

m r r r m r r

01 0 1 0( )

Z r

aR r N e

2 2

1 20 0

1

2(4 ) 1

Z eE

a

2 2 22

2 20

1 ( 1)( ) ( )

2 2

Zr R r ER r

m r r r m r m a r

or

2 2

1 2 20

1

2 1

ZE

ma

or

Note: the alternative forms above have been obtained using:2

00 2

4a

me

02 210

0

Zr

aR Zr N r e

dr a

01 0 1 0( )

Z r

aR r N e

2 2 2

22 2

0

1 ( 1)( ) ( )

2 2

Zr R r ER r

m r r r m r m a r

2 2

1 2 20

1

2 1

ZE

ma

0 02 210

0 0

2Zr Zr

a aR Z Zr N r e e r

r dr a a

0 0

22

10 1020 0

2Zr Zr

a aZ Zr N e rN e

a a

22

20 0

2Z Zr R rR

a a

2 2 22 2

2 2 20 0

1 2

2 2

R Z Zr r R rR

m r r r m r a a

2 2 2

20 02

Z ZR R

ma ma r

01 0 1 0( )

Z r

aR r N e

2 2 2

22 2

0

1 ( 1)( ) ( )

2 2

Zr R r ER r

m r r r m r m a r

2 2

1 2 20

1

2 1

ZE

ma

2 2 2 22

2 20 0

1

2 2

R Z Zr R R

m r r r m a m a r

2 2 2 2 2 2 22

2 2 20 0 0 0

1 ( 1)( ) 0

2 2 2

Z Z Z Zr R r R R R

m r r r m r m a r m a m a r m a r

Therefore:

2 2

202

ZR

ma

1E R

2 2

1 202

ZE

ma

Outline



• Atomic Units




The Hydrogen Atom Wavefunctions

The Complete Wavefunction (Complex Form)

| |, ( ) ( ) ( )n l m n l m l mψ ( r , θ φ ) R r ( ) ( )mi mn l lR r e P

Note that these wavefuntions are complex functions becauseof the term, eim.

This does not create a problem because the probability of findingthe electron in the volume element, dV = r2sin()drdd is given by:

2( , , ) * ( , , ) ( , , ) s i n ( )P r d V r r r d r d *

2( ) ( ) ( ) ( ) s i n ( )m mim imn l l n l lR r e P R r e P r d r d

22 2( ) ( ) s i n ( )m

n l lR r P r d r d

( ) ( , )nl lmR r Y

Real Form of the Wavefunctions

It is common to take the appropriate linear combinations of the complex wavefunctions to obtain real wavefunctions.

This is legal because the energy eigenvalues depend only on n and are independent of l and m; i.e. wavefunctions with differentvalues of l and m are degenerate.

Therefore, any linear combination of wavefunctions with the samevalue of n is also an eigenfunction of the Schrödinger Equation.

p wavefunctions (l = 1)

11 1 1 1 1, ( ) ( ) ( ) s i n ( )i i

n n nψ ( r , θ φ ) R r P e R r e

0 01 0 1 1 1, ( ) ( ) ( ) c o s ( )i

n n nψ ( r , θ φ ) R r P e R r

11 1 1 1 1, ( ) ( ) ( ) s i n ( )i i


zn pψ Already real

p wavefunctions (l = 1) (Cont’d)

1 1

1( ) s in ( ) ( ) s in ( )

2i i

n nR r e R r e 11 1 1

1

2xnp n nψ

1

1( ) s in ( ) co s( ) s in ( ) (co s( ) s in ( ))

2xnp nR r i i

1 1

1( ) s in ( ) ( ) s in ( )

2i i

n nR r e R r ei

11 1 1

1

2ynp n nψi

1

1( ) s in ( ) co s( ) s in ( ) (co s( ) s in ( ))

2ynp nR r i ii

12 ( ) s i n ( ) c o s ( )xn p nR r Real

12 ( ) s i n ( ) s i n ( )yn p nR r Real

p wavefunctions (l = 1) (Cont’d)

12 ( ) s i n ( ) c o s ( )xn p nR r

12 ( ) s i n ( ) s i n ( )yn p nR r

1 ( ) c o s ( )zn p nψ R r

Spherical Polar Coords.

x = rsin()cos()

y = rsin()sin()

z = rcos()

d wavefunctions (l = 2)

12 1 2 2 2, ( ) ( ) ( ) s i n ( ) c o s ( )i i


0 0 22 0 2 2 2, ( ) ( ) ( ) 3 c o s ( ) 1i

n n nψ ( r , θ φ ) R r P e R r

12 1 2 2 2, ( ) ( ) ( ) s i n ( ) c o s ( )i i


2zn dψ Already real

2 2 2 22 2 2 2 2, ( ) ( ) ( ) s i n ( )i i


2 2 2 22 2 2 2 2, ( ) ( ) ( ) s i n ( )i i


d wavefunctions (l = 2) (Cont’d.)

2 2

222 2 2 2

12 ( ) sin ( ) cos(2 )

2x ynd n n nψ R r

2

22 0 2 ( ) 3 c o s ( ) 1

zn d n nψ R r

By the same procedures used above for the p wavefunctions, one finds:

21 2 1 2

12 ( ) sin ( ) cos( ) cos( )

2xznd n n nψ R r

21 2 1 2

12 ( ) sin ( ) cos( ) sin ( )

2yznd n n nψ R ri

222 2 2 2

12 ( ) sin ( ) sin (2 )

2xynd n n nψ R ri

Plotting the Angular Functions

Below are the familiar polar plots of the angular parts of the hydrogenatom wavefunctions.

xy

z

n s

xnp ynpznp

xznd yznd 2znd xynd 2 2x y

nd

Plotting the Radial Functions

R1s has no nodesR2s has 1 nodeR3s has 2 nodes

0 2 4 6 8 10

vals.1

-0.2

0.0

0.2

0.4

r R1S R2S R3S( )

R1s

R2s

R3s

r/ao r/ao

R1s2

R2s2

R3s2

0 2 4 6 8 10

vals.1

0.00

0.02

0.04

0.06

0.08

0.10

r R1S2 R2S2 R3S2( )

R2p has no nodesR3p has 1 nodeR3d has no nodes

r/ao

R2p

R3p

R3d

0 2 4 6 8 10

vals.1

-0.1

0.0

0.1

0.2

0.3

0.4

0.5

r R2P R3P R3D( )

r/ao

R2p2

R3p2

R3d2

0 2 4 6 8 10

vals.1

0.00

0.05

0.10

0.15

r R2P2 R3P2 R3D2( )

In General: (a) A wavefunction has a total of n-1 nodes

(b) There are l angular nodes (e.g. s-0, p-1, d-2)c

(c) The remainder (n-1-l) are radial nodes.

Outline



• Atomic Units




Use of Hydrogen-like Atom Wavefunctions

To illustrate how the hydrogen-like atom wavefunctions may be usedto compute electronic properties, we will use the 2pz (=2p0) wavefunction.

2 2 1 0 ( ) ( ) ( )p z A R r 02

0

cos( )Zr

aZrA e

a

r 0 r < Distance of point from origin (OP)

0 Angle of vector (OP) from z-axis

0 2 Angle of x-y projection (OQ) from x-axis

Review: Spherical Polar Coordinates

2 sin( )dV r dr d d

02

0

cos( )Zr

aZrA e

a

Wavefunction Normalization

* 1dV 0

2

2 2

0

cos( ) sin( )Zr

aZrA e r drd d

a

0

222 4 2

0 0 00

1 cos ( ) sin( )Zr

aZA r e dr d d

a

2

2

0

1 R

ZA I I I

a

0

222 4 2

0 0 00

1 cos ( ) sin( )Zr

aZA r e dr d d

a

2

2

0R

ZA I I I

a

04

0

Zr

aRI r e dr

10

!

nxn nd xex

5

0

4!

Za

5

024a

Z

2

0cos ( ) sin ( )I d

)(c o s

3

1)s in()(c o s 32 xd xxx

3 31 1cos ( ) cos (0)

3 3I

2

0[2 0 ] 2I d

2

3

0

222 4 2

0 0 00

1 cos ( ) sin( )Zr

aZA r e dr d d

a

2

2

0R

ZA I I I

a

5

024R

aI

Z

2

3I 2I

2 52 0

0

21 24 2

3

aZA

a Z

32 0 32

aA

Z

3

2

0

1

32

ZA

a

02

0

cos( )Zr

aZrA e

a

0

3/ 2

2

0 0

1cos( )

32

Zr

aZ Zre

a a

We could have combined the extra Z/a0 into the normalization constant

3/ 2

0

1

32

ZA

a

*r r dV 0

2

2 2

0

cos( ) sin( )Zr

aZrr A e r drd d

a

0

222 5 2

0 0 00

cos ( ) sin( )Zr

aZr A r e dr d d

a

Calculation of <r>

02

0

cos( )Zr

aZrA e

a

3/ 2

0

1

32

ZA

a

2

2

0R

ZA I I I

a

2

3I 2I Same as beforeand

05

0

Zr

aRI r e dr

10

!

nxn nd xex

6

0

5!

Za

6

0120a

Z

6

0120R

aI

Z

2

3I 2I

3/ 2

0

1

32

ZA

a

2

2

0R

Zr A I I I

a

3 2 6

0

0 0

1 2120 2

32 3

aZ Z

a a Z

0 04805

96

a ar

Z Z

Calculation of other Averages

We use the same procedures. For example, I’ll set up the calculationfor the calculation of <y2>, where y = rsin()sin()

2 2*y y dV 0

2

2 2 2

0

sin( ) sin( ) cos( ) sin( )Zr

aZrr A e r drd d

a

0

222 2 6 3 2 2

0 0 00

sin ( ) cos ( ) sin ( )Zr

aZy A r e dr d d

a

2

2

0R

ZA I I I

a

and evaluate.

Outline



• Atomic Units




The Radial Distribution Function

Often, one is interested primarily in properties involving only r, the distance of the electron from the nucleus.

In these cases, it is simpler to integrate over the angles, and .

One can then work with a one dimensional function (of r only), calledthe Radial Distribution Function, which represents the probabilityof finding the electron between r and r+dr.

r/ao

R2p2

R3p2

R3d2

0 2 4 6 8 10

vals.1

0.00

0.05

0.10

0.15

r R2P2 R3P2 R3D2( )

The Radial Distribution Function

Is the most probable value of r for an electron in a hydrogen2p orbital the maximum in R2p

2 ?

No!! Relative values of R2p2 represent the relative probability

of finding an electron at this value of r for a specific pair of values of and .

To obtain the relative probability of finding an electron at a givenvalue of r and any angle, one must integrate * over all valuesof and .

( , , ) ( ) ( ) ( )r A R r P F One can write the wavefunction as:

The probability of finding an electron at a radius r, independent of and is:

2( ) * sin ( )P r dr r drd d

2( ) * s in ( )A R P F A R P F r drd d

22 2

0 0( ) ( ) * ( ) * sin ( ) *P r dr A R r R r r dr P P d F Fd

where22

0 0* sin( ) *B A P P d F Fd

i.e. we’ve incorporated the integrals over and into B

22( ) ( )P r d r B r R r d ror

22( ) ( )P r B r R r is called the Radial Distribution Function (orRadial Probability Density in this text)

We’ll use RDF(r) P(r)

RDF1s has 1 maximum RDF2s has 2 maximaRDF3s has 3 maxima

0 2 4 6 8 10

vals.1

0.0

0.1

0.2

0.3

r RD1S2 RD2S2 RD3S2( )

RDF1s

RDF2s

RDF3s

r/a0

R1s has no nodesR2s has 1 nodeR3s has 2 nodes

This is because:

We’ll use RDF(r) P(r)

RDF1s has 1 maximum RDF2p has 1 maximumRDF3d has 1 maximum

0 2 4 6 8 10

vals.1

0.0

0.1

0.2

0.3

r RD1S2 RD2P2 RD3D2( )

RDF1s

RDF2p

RDF3d

r/a0

R1s has no nodesR2p has no nodesR3d has no nodes

This is because:

Most Probable Value of r

0 2 4 6 8 10

vals.1

0.0

0.1

0.2

0.3


RDF1s

RDF2p

RDF3d

r/a0

The most probable value of the distance from the nucleus, r, isgiven by the maximum in the Radial Distribution Function, P(r) RDF(r)

It can be computed easily by: 22 ( )( )0

d Br R rdP r

dr dr

The wavefunction for an electron in a 2pz orbital of a hydrogenlike atom is: 02

2 cos( )z

Zr

ap Are

We will determine the most probable distance of the electronfrom thes nucleus, rmp.

0

2

2 22 2( )Zr

aP r Br R Br re

04

Zr

aBr e

04 0Zr

amp

dBr e when r r

dr

0 0

440

Zr Zr

a ad drB r e e

dr dr

0 04 3

0

4Zr Zr

a aZB r e e r

a

One gets the same result for any 2p orbital because theRadial portion of the wavefunction does not depend on m.

0 04 3

0

0 4Zr Zr

a aZB r e e r

a

03

0

4Zr

a ZrBr e

a

04mp

ar

Z

Therefore:0

4 0mpZr

a

0 2 4 6 8 10

vals.1

0.0

0.1

0.2

0.3


RDF1s

RDF2p

RDF3d

r/a0

By the same method, one may calculate rmp for 1s and 3d electrons:

04(2 )mp

ar p

Z

0(1 )mp

ar s

Z

09(3 )mp

ar d

Z

These most probable distances correspond topredicted radii for the Bohr orbits:

2 0n

ar n

Z

Probability of r in a certain range

10

!n axn

nx e dx

a

Some NumericalIntegrals

3 4

04.43xx e dx

5 4

013.43xx e dx

We will again consider anelectron in a 2p orbital, for which: 04( )

Zr

aP r Br e

What is the probability that 0 r 5a0/Z

We will first normalize the RDF

04

0 01 ( )

Zr

aP r dr B r e dr

5

0

4!B

Za

5

024a

BZ

5

0

1

24

ZB

a

10

!n axn

nx e dx

a


3 4

04.43xx e dx

5 4

013.43xx e dx

04( )Zr

aP r Br e

What is the probability that 0 r 5a0/Z

5

0

1

24

ZB

a

05 /

0 0(0 5 / ) ( )

a ZP r a Z P r dr 0

05 / 4

0

Zra Z aB r e dr

Define:

0

Zrx

a Then: 0a

r xZ

0adr dx

Z 0

0

5 5a Z

r x rZ a

45

0 00 0

(0 5 / ) xa aP r a Z B x e dx

Z Z

5

5 40

0

xaB x e dx

Z

5 5

0

0

113.43

24

aZ

a Z

13.43

24 0 .5 6

What is the probability that 3a0/Z r < 10

!n axn

nx e dx

a


3 4

04.43xx e dx

5 4

013.43xx e dx

04( )Zr

aP r Br e

5

0

1

24

ZB

a

One can use the identical method used above to determinethat:

0

4.43(0 3 / ) 0.18

24P r a Z

0 0(3 / ) 1 (0 3 / )P a Z r P r a Z

0 0(3 / ) 1 (0 3 / )P a Z r P r a Z 1 0 .1 8 0 .8 2

Calculate <r> for an electron in a 2pz orbital

(same as worked earlier using the complete wavefunction)

10

!n axn

nx e dx

a

04( )

Zr

aP r Br e

5

0

1

24

ZB

a

0( )r rP r dr

04

0

Zr

ar Br e dr

05

0

Zr

aB r e dr

5

60

0

1 5!

24

Zr

a Za

1

0

1 120

24 Za

05a

Z Same result as before.

10

!n axn

nx e dx

a

04( )

Zr

aP r Br e

Calculate the average potential energy for an electron in a 2pz orbital.

5

0

1

24

ZB

a

0

1 1( )P r dr

r r

04

0

1Zr

aBr e drr

03

0

Zr

aB r e dr

5

40

0

1 1 3!

24

Z

r a Za

0

16

24

Z

a

0

1( )

4

Ze eV r

r

2

0 0

1 1

4 4

Ze e ZeV

r r

0

1

4

Z

a

2 2

0 0 0

1 1

4 4 4

Ze Ze ZV

r a

2 2

0 016

Z e

a

2 2

0 016

Z eV

a for an electron in a 2p orbital

2 2

20 0

1

2(4 )n

Z eE

a n Total Energy:

2 2

20 032

Z eE

a

Note that <V> = 2•E

Calculation of average kinetic energy, <T>

T V E

2T E E

T E

Also: 2 2V E T

Signs

<V> negative<T> positive<E> negative

Outline



• Atomic Units




Atomic Units

All of these funny symbols flying around are giving me a headache.Let’s get rid of some of them.

Let’s redefinesome basic units: me=1 (mass of electron)

e = 1 (charge of electron)

ħ = 1 (angular momentum)

4o = 1 (dielectric permittivity)

Derived Units2

00 2

4

e

am e

Length: 1 au 1 bohr

Energy:2

10 0

2 ( ) 1 hartree (h)4s

eE H

a

Conversions

1 bohr = 0.529 Å

1 h = 2625 kJ/mol

1 h = 27.21 eV

Hydrogen Atom Schrödinger Equation

2 22

02 4e

ZeE

m r

2 4

2 2 20

1

2(4 )e

n

m Z eE

n

2 4

1 2 202(4 )

em Z eE

21

2

ZE

r

SI Units Atomic Units

2

2

1

2n

ZE

n

2

1 2

ZE

slide 1 chapter 6 the hydrogen atom. slide 2 outline the hydrogen atom schrödinger equation the...

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hydrogen atom slide

rcos slide

contd real slide

radial wavefunctions

ze e r slide

real wavefunctions

complex wavefunctions

p wavefunctions