Chapter 11 - THE QUANTUM DEFECT
Problems with solutions
11.1 Using quantum defects calculate the ionization potentials of lithium and sodium in
electron volts. The ionization potential is simply the ground state energy of the atom.
Solution:
Since the electron configurations of the ground states are 1s 2 2 s and 1s 2 2s 2 2 p 6 3s we
require only the quantum defects of the s-states of each atom. For Li and Na they are,
respectively 0.40 and 1.35. Therefore
13.6 eV
13.6 eV
IP  Li  
 5.31eV ; IP  Na  
 5.00 eV
2
2
 2  0.4 
 3  1.35 
11.2 A spherically symmetric singly charged positive ion (the "ionic core") is situated at the
origin of coordinates. An electron is assumed to be at a fixed distance z along the z-axis. This
electron induces in the ion a dipole moment, the magnitude of which is determined by the ion's
polarizability  d . Assume that z  the dimension of the ion.
a) Using the result from electromagnetic theory that the field of a dipole is given by
p
Edipole  r , ,   
2 cos  rˆ  sin  θˆ find the force on the electron.
4 0 r 3
b) Show that the work done in bringing the electron from infinity to z  r gives the dipole
potential energy in Equation (11.8) after, of course, converting from the SI units of this problem
to atomic units.
Solution:
The magnitude of the dipole moment p is given by p   d Ee where Ee is the electric field of the
electron at the origin. Therefore
e 1
p  d
so the electric field of the dipole at a point on the z-axis is
4 0 z 2


2
 1 
1 
e 11
1
Edipole  z  
2 kˆ  
 d
 2e d 5 kˆ
2  3  
4 0  4 0 z  z
z
 4 0 
2
 e 
1
the force on the electron is thus F  eEdipole kˆ   
 2 d 5 kˆ which is, as expected,
z
 4 0 
attractive.
b) The work done in bringing the electron to z  r from infinity is the potential energy so
2
V
d
pol
r 1
 e 
 r    F  dr   
 2 d  5 kˆ   dzkˆ
z
 4 0 

r
2


2
 e 
 e  d
1 1 

 2 d  4    

4
4  z r
 4 0 
 4 0  2r
d
When converted to atomic units we obtain V pol
r   
d
2r 4
.
11.3 The effects of core penetration by the valence electron of a Rydberg atom may be
approximated by assuming that the potential as seen by the Rydberg electron is, in atomic units,
1 b 
V  r      2  where b is a positive constant. The effect of this potential is that of increasing
r r 
Solution to Chapter 11 problems
page 1
Chapter 11 - THE QUANTUM DEFECT
Problems with solutions
the attraction between the valence electron and the nucleus (of charge Z) as the Rydberg electron
penetrates the ionic core. The effective nuclear charge Z' and b are then chosen to give the best
fit to spectral data. By averaging the 1/ r 2 term in the potential energy over a Keplerian orbit
find the quantum defect in terms of the Z', b and the angular momentum.
Solution:
Use the integral
m2
 1  1 2 2 m 2
r m  
     1   cos   d
3 
0
 2 n 
1
b
to obtain for m  2 r 2  3 so the average value of the b / r 2 term in the potential is 3 .
n
n

Comparing with En  En  3 we see that   b / .
n
1 b 
11.4 The quantum defect for the potential of the previous problem, V  r      2  where b is
r r 
a positive constant, may also be obtained from the quantum mechanical solution of the radial part
of the Schrödinger equation because the 1/ r 2 term may be combined with the centrifugal term.
Show that the quantum defect and is given by   b /   1/ 2 . Can you account for the
difference between this answer and the previous answer? How do these energy states vary for a
given n with ?
Solution:
The effective potential is Veff r   V r  
  1
. Since the potential in our problem contains a
2r 2
1/r2 term we may include it with the centrifugal term in the effective potential. We may then
write the effective potential for our problem in the form
  1   1  b    1   1     1  2b
Veff  r   V  r  
2r 2
r r2
2r 2
r
2r 2
The expression in the square brackets replaces (+1) in the "pure" hydrogen atom problem so let
1
where
'  ' 1    1  2 . In the solution for the hydrogen atom we had n  n '  1 
2E
n' is the number of terms after which the finite series terminates in order to keep it from
diverging. The energy in the pure hydrogen atom problem is given by
1
1
E 
. In our problem  n '  1  n*   n    , the effective quantum number.
2  n '  12
That is, ' replaces. We have then En  
1
2  n ' ' 1
2

1
2n  

2
. Now we have to find
 . Adding and subtracting , we have n*  n'  1  '  n    '  n    so that
     ' . Now, we re-write the above equation '  ' 1 
'2  '  2b 


 1  2b as
 1   0 and solve for ' using the quadratic formula.
Solution to Chapter 11 problems
page 2
Chapter 11 - THE QUANTUM DEFECT
Problems with solutions
2
1
1
1

   2b        2b

4
2
2

2
2
 1

1
1
1



Then    '          2b           2b
2
2
2
 2
 




1 1
' 
1  8b  4
2 2
1
 1   
2
2
It is immediately clear that it is the plus sign that must be discarded because, in the absence of the
1/ r 2 term, which corresponds to b  0 , the quantum defect must vanish, i.e. it is a pure hydrogen
atom. We have then
1/ 2
 
 
 
2
 
1
1
1 
2b  



          2b     1  1 

2
2
2
2  



1  
  
  
2   





1
b
b

  
... 
2
1
2  

1  
      2 

2  

The answer to the previous problem is   b / . The   1/ 2 clearly comes from the
term in the solution of the radial Schrödinger equation, a quantum mechanical equation.
The energy is




1
1 
1
2b 



En  


1

2

...


1



2
1 
2n 2 
n
2n 2 


2n  
n  

2  


 1
11.5 In addition to the dipole term in the Hamiltonian that led to the quantum defect
3
 d   d  5  there is also a potential energy due to quadrupole polarizability of the ionic core
4
q
that can be included. This quadrupole term is V pol
r    q / 2r 6 . Use the same technique as
that used in the text to show that the portion of the quantum defect that is due to quadrupole
35
polarizability is  q   q  9  .
16
Solution:
q
r   
The potential is given by V pol
q
6
so we must evaluate r 6 . The general result derived
2r
in the text is

 32 m  m  2m  3 2  1  m  2m  3m  4m  5 4  3 
r m  3 1 
  
    ...
2!
4!
n 
 2
8

which may be evaluated for m = 6. We have
Solution to Chapter 11 problems
page 3
Chapter 11 - THE QUANTUM DEFECT
Problems with solutions
 4  3 2  1  4  3  2 1 4  3 

 1 
r 6   3    9 1 
  
    ...
2!
4!
n 
 2
8


  9  
3 
  3   1  3 2   4 
8 
n  
2
Now  2  1  2 , but we neglect terms of higher order in  2 / n 2 so   1 and we have
n
9
  
3
r 6   3   1  3  
8
n  
  9   35 
  3    
n   8
Then,
q
r   
V pol

q
2

1
r6
 q   9   35 



 
2  n 3   8 
35

 q  9
16


Solution to Chapter 11 problems
page 4