1
Hydrogenic Atoms
Introduction
Hydrogenic atoms are atoms with nucleus (H+, Fe26+, Pb82+, etc…) and one electron.
The hydrogenic atom has an analytic solution.
- I.e., the solution is exact, no approximations needed.
Coulomb Potential
From Coulomb’s law, the potential energy between any two charges, q1 and q2, is
1 q1q 2
V r  
40 r12
- r12 is the distance between the two charges
- 0 is the permittivity of free space
For a hydrogenic atom, the potential energy can be written as
1 Ze2
V r  
40 r
- r is the distance of the electron from the nucleus.
- Z is the charge of the nucleus
- e is the fundamental unit of charge (i.e., the charge of e-)
Center of Mass and Relative Coordinates
Hydrogenic atoms have two particles; therefore, the Hamiltonian can be written as
H
p12
p2
 2  V r 
2m1 2m 2
At this point we note that the potential energy does not depend on where the total system is in
space. However, the potential energy does depend on the position of the particles relative to
each other.
The simplest coordinate system one could choose is based on the Cartesian coordinates of
each particle, so that, H is a function of x1, y1, z1, x2, y2, z2.
2
This simple coordinate system can be transformed into a system described by center-of-mass
coordinates and relative coordinates.
z2
x2
z
y2
z1
x1
y1
z12
x12
ZCM
XCM
r12
YCM
y12
r2
r1
z
RCM
x
x
y
y
Thus the Hamiltonian can be rearranged in terms of the center-of-mass and relative
coordinates.
2
pCM
p2
 rel  V  r12 
2M 2
m1m 2
1 1
1

where M  m1  m2 and  
or 
m1  m 2
 m1 m 2
 is known as the reduced mass of the system.
H
Note for the hydrogen atom
1
1
1
1
1




4
 me mp 5.48580 10 amu 1.00727 amu
   5.48281104 amu
3
Schrödinger Equation
The coulombic potential is radially symmetric, that is, the value of the electric field coming
from the nucleus depends only on how far away from the nucleus we are. The value has no
dependence on the orientation, i.e., the angular variables. Thus using a radially symmetric
coordinate system such as the polar spherical coordinates system would be sensible.
The kinetic energy operator in polar spherical coordinates is
T̂ 
 2
2
1   2  
1
 
 
1
2 

sin


 2 r

 2


  r 2 sin 2  2 
 r r  r  r sin   
Note the relationship between the kinetic operator in polar spherical coordinates and the 3-D
angular momentum operator.
1   2  
1
 
 
1
2 
 2 r

 2
 sin    2 2
  r sin  2 
 r r  r  r sin   
 2  1   2   1  1  
 
1  2    2  1   2  
1 ˆ2 

r

sin



 2 
 2 r
 2


  2 2  L  
2
2 
2  r r  r  r  sin   
  sin     2  r r  r 
r

 2  1   2   L̂2 


r
 2 2 
2  r 2 r  r 
r 
1 Ze2
Recall that the coulombic potential is V  r   
40 r
Thus, the Hamiltonian and the Schrödinger equation can be written as
 2
T̂ 
2
2
1 Ze 2
 1   2   Lˆ 
r


 2 

2 2
r  4 0 r
 r r  r 
 2  1   2   Lˆ 2  
1 Ze 2
Hˆ  
r


  E



2 2
2  r 2 r  r 
r  4 0 r
To solve this partial differential equation, we will find success using the separation of
variables technique.
2
ˆ 
Hˆ  Tˆ  V
2
Let   r, ,   R  r       
 2
2
2
 1   2 

 Lˆ
R  r           2 2 R  r         
 2 r
r

 r r  r

1 Ze 2

R  r         E R  r       0
40 r
4
Allow the differential operators to operate.
 2
2
           2 R  r  
Lˆ 2        


r
  R r
2 2
r2
r 
r 
r



Recall that L̂2        
 2
2
2
1 Ze2
R  r         E R  r         0
40 r
l  l  1        
           2 R  r  
l  l  1 
r

R
r














r2
r 
r 
r 2 


1 Ze 2
R  r         E R  r       0
40 r
Now divide the equation by   r, ,   R  r       
 2
2
1 Ze2
 1 1   2 R  r   l  l  1 
r


E 0




2
r 
r 2  40 r
 R  r  r r 
 2   2 R  r  
r

R  r  2r 2 r 
r 
1
l  l  1
1 Ze2

E 0
2r 2
40 r
2
Radial Equation
Thus the radial equation for a hydrogenic atom is
2

 2   2 R  r    l  l  1
1 Ze 2
r


 E R r   0



2
2
2r r 
r   2r
4 0 r

The equation can be related to a differential equation called the generalized Laguerre
equation.
The solutions to the generalized Laguerre equation depend on two quantum numbers, n and l.
Zr
 2Zr 
1   n  l  1!  2Zr   na0
R n,l  r   

 e Ln l,2l1 

n  a 0  n  l !   na 0 
 a0 
l 1
10 2 2
1 3
34
40 2 1.112650 10 C s kg m 1.05459 10 J  s 
a0 

2
me e2
 9.10953 1031 kg 1.60219 1019 C 
 5.29177 1011 m  0.529177Å = 52.9177 pm
2
5
 2Zr 
The Ln l,2l1 
 are called the generalized Laguerre polynomials.
a
 0 
A Table of Laguerre polynomials.
n
l
Ln+1, 2l+1
1
0
L1,1(x) = -1
2
0
L2,1(x) = -2! (2 - x)
2
1
L3,3(x) = -3!
3
0
L3,1(x) = -3! (3 - 3x + ½ x2)
3
1
L4,3(x) = -4! (4 - x)
3
2
L5,5(x) = -5!
4
0
L4,1(x) = -4! (4 - 6x - 2x2 - 1/6 x3)
4
1
L5,3(x) = -5! (10 - 5x + ½ x2)
4
2
L6,5(x) = -6! (6 - x)
4
3
L7,7(x) = -7!
x = 2Zr/a0
x = Zr/a0
x = Zr/a0
x = 2Zr/3a0
x = 2Zr/3a0
x = 2Zr/3a0
x = Zr/2a0
x = Zr/2a0
x = Zr/2a0
x = Zr/2a0
A Table of Radial Wavefunctions,  n,l
3
2
3
2
1,0
Z
 2  e
 a0 
3,0
2
Zr
 Z  2   3a 0
2  Z 2 
Z

   27  18 r  2   r  e
a0
81 3  a 0  
 a0  

Zr
a0
 2,0
1 Z 
Zr 

  2
e
a0 
2 2  a0  

Zr
2a 0
5
2

1 Z
2a 0
 2,1 
  re
2 6  a0 
3
7
5
2
Zr
 Z  2   3a 0
4  Z 2  Z
 3,1 
   6 r    r  e
81 6  a 0   a 0
 a0  
3,2
4  Z  2 3  3a 0

  r e
81 30  a 0 
Zr
3
 4,0
2
3
Zr
 Z 2 
 Z  2  Z  3   4a 0
Z
 6   192  144 r  24   r    r  e
a0
 a 0  
 a0 
 a 0  
5
2
Zr
 Z 2 
1
Z 2  Z  3   4a 0
 4,1 
   80 r  20 r    r  e
a0
256 15  a 0  
 a0  
7
 4,2
1  Z 2 
Z 3   4a 0
2

12
r

r e
  
a0 
768 15  a 0  
9
Zr
 4,3
 Z  2 4  4a 0
4

  r e
768 35  a 0 
Zr
Zr
6
Plots of Radial Wavefunctions
0.559
0.5
0.4
1s
w 1s( r) 0.3
w 2s( r)
y 3s( r) 0.2
2s
3s
0.1
0
 0.027 0.1
0
5
10
15
0.01
20
r
25
25
0.08
0.06
3p
2p
y2p( r)
0.04
3d
y3p( r)
y3d( r)
0.02
0
0.02
0
5
10
15
r
Note: s wavefunctions are non-zero at nucleus.
p, d, f, … are zero at nucleus (for point nucleus).
20
25
7
Radial Distribution Function
0.043
0.05
1s
0.04
a( r)
0.03
b( r)
2s
d( r) 0.02
3s
0.01
0
0
0
5
10
0
0.047
15
20
r
25
25
0.05
2p
0.04
3d
3p
c( r) 0.03
e( r)
f ( r) 0.02
0.01
0
0
0
0
5
10
15
r
Note: Where radial probabilities are zero, radial node where Ln,l  x   0
Number of radial node = n – (l + 1)
20
25
25
8
Hydrogenic Wavefunctions
Recall that the hydrogenic wavefunction is a product of the radial wavefunction and the
spherical harmonic. n,l,ml  r, ,   R n,l  r  Y
1s
2s
2p
3p
4p
3s
3d
4d
Note: orbital not to scale
Note: l – indicates number of angular nodes
5p
9
Introduction to Electronic Spectra
Spectrum is measured by observing how light interacts with matter, creating excitations and
deexcitations.
For electronic spectra,
- the values the energy levels are of secondary importance
- the difference in energy levels is of primary importance
Energy of a hydrogenic atom
En  
Z2e4
1
1
 2  R H  2
2 2 2
32 0
n
n
Spectral lines for a hydrogenic atom
E 
Z2e4  1 1 
  
32202 2  n i2 n f2 
Selection rules – the allowed changes in quantum numbers
– i.e., allowed transitions between energy levels.
Change of the principal quantum number
n  1, 2, 3,
Change of the azimuthal quantum number
l  1
Change of the magnetic quantum number
ml  0, 1
The l  1 selection rule implies that the 2s  1s transition is a forbidden transition
10
Energy level diagram a hydrogenic atom
4s
3s
4p
3p
2s
2p
4f
4d
3d
1s
Note: For a hydrogenic atom, the energy levels depend only on the principal quantum
number. For a multi-electron atom (i.e., two or more electrons), the energy levels depend on
both the principal quantum number and the azimuthal quantum number.
Degeneracy of atomic orbitals
Degeneracy of p orbitals
The wavefunctions for p orbitals are in terms of the spherical harmonics
1
 3 2
Y10  ,      cos 
 4 
1
1
 3 2
 3 2
Y11  ,      sin  ei Y11  ,      sin  ei
 8 
 8 
Note that Y1  ,  and Y11  ,  are complex functions.
1
We are not able to visualize a complex orbital. However, since Y1  ,  and Y11  ,  are
degenerate, any linear combination of them is also a solution to the Schrödinger equation.
1
We can make real wavefunction by taking the following linear combinations
1
1
1 1
 3 2
 3 2
i
 i
px 
Y1  Y11   
sin

e

e





  sin  cos 
2
 16 
 4 
1
1
i 1
 3 2
 3 2
i
 i
py 
Y1  Y11   i 

 sin   e  e     sin  sin 
2
 16 
 4 
There is no substantial difference between the Y1  ,  and Y11  ,  and the px  , 
1
and p y  ,  orbitals. Choosing one set over another is matter of convenience.
11
Degeneracy of d orbitals
1
 5 2
2
Y  ,    
  3cos   1
 16 
0
2
1
 15  2
Y21  ,      sin  cos ei
 8 
1
1
 15  2
Y21  ,      sin  cos ei
 8 
1
 15  2 2 2i
Y22  ,    
 sin e
 32 
 15  2 2 2i
Y22  ,    
 sin e
 32 
After linear combination
1
 5 2
2
d z2  ,    Y  ,    
  3cos   1
 16 
0
2
d xz 
1
2
1 1
 15 
Y2  Y21     sin  cos  cos 

2
 4 
1
2
d yz 
i 1
 15 
Y2  Y21     sin  cos  sin 

2
 4 
d x 2  y2
1
 15  2 2

Y22  Y22   

 sin  cos 2
2
 16 
1
1
i 2
 15  2 2
d xy 
Y2  Y22   

 sin  sin 2
2
 16 