9-1
Chapter 9 The Hydrogen Atom
Goal - to solve for all eigenstates (orbitals)
H atom - single nucleus, charge Z (+1) and one eattracted by Coulomb’s Law
Will find third quantum number, n, that is ≥ 1
1. Write the full Hamiltonian. Now need to let r
vary since real atoms don’t have fixed distances
between nuclei and electrons.
2. Use separation of variables to pull out part
already solved for rigid rotor (angular part).
3. Show solution for radial part (r dependent part).
E energy will depend only on the n quantum
number (and not m or l).
Finally, we draw pictures of the orbitals and count
up the orbital nodes and level degeneracy.
Hamiltonian is the sum of the kinetic energy of
each particle plus the potential energy, which here
is Coulombic attraction. (Recall Coulomb’s Law)
0 = permittivity
constant
Distance between charges
9-2
The Schrodinger equation in polar coordinates:
 1   2  (r ,  ,  ) 
1
 
 (r ,  ,  )  
r

sin


 2

 
2  2
r

  r r 
e2
 r sin   
 


 (r ,  ,  )
2

2me  1
4

r
  (r ,  ,  )
0
 2

 2
 r sin 

 E (r ,  ,  )
Radial Equation: Separation of Variables
Notice the potential only depends upon r.
That means it’s separate from the  and  parts.
The total wave function then can be written as a
product of the part we did for angular momentum
and the radial part:
This allows us to get the radial only equation.
 (r ,  ,  )  R(r )( )( )
We already did the  and  parts to generate the
spherical harmonics.
9-3
We can make substitution recalling that
lˆ( )( )   2l (l  1)( ) ( )
To find:
 2 d  2 dR(r )    2 l (l  1)
e2 

r


 R(r )  ER(r )
2
2


dr   2me r
40 r 
2me r dr 
Second term can be viewed as an effective
potential. (Centripetal plus Coulombic parts)
9-4
Recall, for the harmonic oscillator, the fundamental
solution was a Hermite polynomial, here now the
fundamental part is an exponential. See page164.
Z is the charge on the nucleus.
 0h2
a0 
me e 2
This is the Bohr radius.
An,l is a constant
Total energy eigenvalues are negative by
convention. Zero is taken at r ∞. Negative
energies correspond to bound states.
9-5
Potential is drawn as before.
Yellow is classically forbidden region.
Energy eigenvalues are superimposed.
Notice that they get closer
as n  ∞.
Need 3 Quantum numbers to describe state of H
9-6
Complete normalized total energy eigenfunctions
include spherical harmonics.
Note that EF real only if ml = 0!
Convenient to linearly combine orbital functions
with their complex conjugate to create real
functions.
9-7
Absorption spectrum of Hydrogen
1s orbital
9-8
“Leading edge” of
contour plot for
s functions is R(r)
Note nodes in R(r)
9-9
To display H total energy
EF other than ns,
need contour plot of
.(x,y,z)with
x or y or z = 0.
Note different
appearance of angular and
radial nodes.
Clockwise from top left: 2py, 3py, 3dz , 3dxy
2
3-d representations
9-10
Calculating the probability of finding the ewithin volume element dV is proportional to 2
Total energy EF have n-l-1 radial and l angular
nodes
Example Problem 9.3 Locate the nodal surfaces in
The angular part, cos, is zero for  =/2. In 3D
9-11
The radial part of the equations is zero for finite
values of
for
This occurs at r= 0, and at r= 6 a0. The first value is
a point in three dimensional space and the second
is a spherical surface.
In an atom, the probability of finding e-at certain
distance, r, from the nucleus is of more interest
than finding it in dV.
Define radial probability
distribution function (rpd)
as P(r)dr
P(r)dr gives the probability of finding the electron
in a spherical shell of radius rand thickness dr.
9-12
Example for 1s orbital. Note that rpd function goes
to zero as r goes to zero
9-13
In Quantum mechanics we draw probability
density map instead of the “planetary” model
9-14
Example Problem 9.6
Calculate the maxima in the radial probability
distribution for the 2s orbital. What is the most
probable distance from the nucleus for an electron
in this orbital? Are there subsidiary maxima?
Plot P(r) and
Principal maximum in P(r)is at 5.24a0.
Subsidiary maximum is at 0.76a0.