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Transcript of Slide 1 Chapter 6 The Hydrogen Atom. Slide 2 Outline The Hydrogen Atom Schrödinger Equation The...
Slide 1
Chapter 6
The Hydrogen Atom
Slide 2
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
Slide 3
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
Slide 4
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
Slide 5
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:^
Slide 6
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 ^
Slide 7
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.
Slide 8
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
Slide 9
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
Slide 10
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
Slide 11
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
Slide 12
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
Slide 13
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
Slide 14
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
Slide 15
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
Slide 16
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
Slide 17
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
n n nψ ( r , θ φ ) R r P e R r e
zn pψ Already real
Slide 18
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
Slide 19
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()
Slide 20
d wavefunctions (l = 2)
12 1 2 2 2, ( ) ( ) ( ) s i n ( ) c o s ( )i i
n n nψ ( r , θ φ ) R r P e R r e
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
n n nψ ( r , θ φ ) R r P e R r e
2zn dψ Already real
2 2 2 22 2 2 2 2, ( ) ( ) ( ) s i n ( )i i
n n nψ ( r , θ φ ) R r P e R r e
2 2 2 22 2 2 2 2, ( ) ( ) ( ) s i n ( )i i
n n nψ ( r , θ φ ) R r P e R r e
Slide 21
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
Slide 22
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
Slide 23
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( )
Slide 24
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.
Slide 25
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
Slide 26
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
Slide 27
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
Slide 28
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
Slide 29
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
Slide 30
*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
Slide 31
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
Slide 32
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.
Slide 33
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
Slide 34
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.
Slide 35
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 .
Slide 36
( , , ) ( ) ( ) ( )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)
Slide 37
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:
Slide 38
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:
Slide 39
Most Probable Value of r
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
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
Slide 40
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
Slide 41
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
Slide 42
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
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
Slide 43
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
Slide 44
10
!n axn
nx e dx
a
Some NumericalIntegrals
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
Slide 45
What is the probability that 3a0/Z r < 10
!n axn
nx e dx
a
Some NumericalIntegrals
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
Slide 46
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.
Slide 47
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
Slide 48
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
Slide 49
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
Slide 50
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
Slide 51
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