Download The potential energy outside the nucleus is

The potential energy outside the nucleus is:
kq2
U r   
r
r  r0
Where k is the Coloumb constant.
The charge inside the sphere as a function of the radius f the sphere is (from volume
ratios):
qr 3
qi  3
r0
Thus the electric field is:
E
kqi kqr
 3
r2
r0
And the potential is:




kq r02  r 2
kq r02  r 2
kq 0
kq


V r    E r dr  3  rdr 

V
r


0
3
3
r0
r0 r
2r0
2r0
r
r0
r
Thus the potential energy is:
kq2
U r   qV r  
2r0
 r  2

   3
 r0 

The Hamiltonian of the unperturbed hydrogen atom is:
p 2 kq2
H0 

2
r
And the new Hamiltonian is:
p 2 kq2
H

2 2r0
 r  2

   3
 r0 

So we can rewrite the perturbed Hamiltonian as:
H  H0  H 
H 
kq2 r 2 3kq2 kq2


2r0
r
2r03
So all that is left is to calculate the perturbation correction for the ground wave function:
100  1,0,0  R10Y00
Because the ground-state wave-function has only radial dependence we get:
E01   100 H   100
r0
E0   Y00  ,   d  R10 r  H r r 2 dr
1
2
2
0
But normalization requires:
 Y  ,  d  1
2
00
And:
R10 r  
R10 r0  

2
e
a03
r
a0

2
a03
e
r0
a0
But since r0  a0 we can conclude that the radial wave function is practically constant
for r<r0:
R10 r0  
2
a 03

e
r0
a0

2
a 03
So the integral simply becomes:
r0

R10 r  H r r 2 dr 
2
0
4kq 2
a 03
r
4 0  kq 2 r 2 3kq 2 kq 2  2



 r dr 
2r0
r 
a 03 0  2r03
r
r0
 1 r0 4
 4kq 2  1 r05
3 0 2
3 r03 r02 
r
dr

r
dr

rdr


 3
0  a03  2r03 5 2r0 3  2  
2r0 0
 2r0 0

4kq 2
 3
a0
 r02 r02 r02  2q 2 r02
 
 
2
5a03
10 2
Thus the correction is:
r0
E0   R10 r 
1
0
2
2kq2 r02
H r r dr 
5a03
2
But the ground-state energy of te unperturbed atom is:
E0 
kq 2
 13.6eV
2a 0
Hence:
2
1
E0
2
 10 15

2q 2 r02 4  r0 
4



  3.9  10 9 eV


E


13
.
6

0
3
10 

5  a0 
5
5a 0
0
.
53

10


For a hollow spherical shell, the calculation becomes much simpler. Retracing our steps
we see that inside the shell the total charge is zero. Hence, from Gauss’ law, the electric
field is zero and the potential is constant. The potential inside the shell is equal to the
potential on the surface of the shell:
kq
V 
r0
So the potential energy inside is:
U  qV  
kq2
r0
Thus the Hamiltonian inside is:
H
p 2 kq 2

2
r0
Comparing this to the unperturbed Hamiltonian:
p 2 kq2
H0 

2
r
We see that the perturbation is:
H  H0  H 
1 1 
H   kq2   
 r r0 
So the correction now is (again, only the radial integral counts and we simplify the
ground-state wave function):
r0
E 0   R10 r 
1
0
2
H r r 2 dr 
4kq 2 0  1 1  2
   r dr 
a 03 0  r r0 
r
r
4kq 2 0  r 2 
4kq 2  r 2 r 2  4kq 2 r 2
 3  r   dr  3  0  0   3 0
r0 
3
a0 0 
a0  2
a0 6
And:
kq 2
E0 
 13.6eV
2a 0
Thus:
2
1
E0
2
 10 15

8  r0 
8
  6.5  10 9 eV
 E 0     13.6  
10 
6  a0 
6
 0.53  10 
The ratio of the muon’s mass to the electrons mass is:
MeV
m
c 2  214

MeV
me
0.51 2
c
Bohr’s radius for the hydrogen is given by:
109
a0 
2
kme q n q e
Where qn and qe are the charge of the nucleus and the electron respectively (in the
hydrogen atom they are obviously the same).
Thus for a lead nucleus – muon system, the modified Bohr radius (that effects the
ground-state wave function) is:
a0 
2
k 214me 82qe qe

O
a0
6
6

57

10
a

30

10
A
0
17.5  10 3
On the other hand, the radius of the nucleus is given by:
r0  A
1
3
O
  12  10 6 A
O
r0  71  10 6 A
Therefore the spherical shell radius is twice as large as the orbital radius of the ground
state.
Thus our assumption that the wave function inside the nucleus is constant is no longer
valid and we need to take into account the exponential term as well.
R10 r  
2
a 0 3

e
r
a0
So the integral for the perturbation is:
r0
E 0   R10 r 
1
0
2r
4kq 2 0  1 1   a0 2
H r r dr 
   e r dr
a 0 3 0  r r0 
r
2
2
Which simplifies:
1
E0

4k 82q 2

a0 3
 
r0
 re
 0

2r
a0
2r
r 2  a0 
dr   e dr 
r

0 0
r0
Integrating by parts (or using integral tables) yields:
x
n
e x dx 
1

x n ex 
n
x

n 1 x
e dx
Thus:
r0
 re

2r
a0
0
r0
2r
2r
2r
2r
r


 

a0  0  a0
a
 a0  a0  a0
a0
a 
dr 
  e dr  re     e  re 0     0
2r  0
2r
2r
4
 
 
 
0
2r
2r

 0
 0
  a   2r e a0  ae a0
0
 0

And by the same token:
a0  a
r 2  a0
0 r0 e dr  4r0 e 0
r0
2 r0
2r
 2 2ar0

 a  e 0  2r 2  2a  r  a  2 
0
0
0 0
0




So together:
1
E0




2r
2r
2r
r
r0 2  2 r
 0
 0
 4k 82q 2 a  2 
4k 82q 2  0  a0
r
a0
a0
0 

r0  r0 e
 a0  a0 e a0
  re dr   e dr  
3
3
r
r0 
a0
a0
 0

0 0
Using the fact that a0  0.36r0 the term in the exponential is a very negative number,
thus the exponential terms disappear leaving us with:
E01 
4k 82q 2 
 0.16a0 2
a03
E01  1.28 
k 82q 2 
2a0








The ground state energy of this system (analogous to the hydrogen atom) is:
E0 
k 82q 2 
2a0
Thus the correction due to the perturbation is larger than the unperturbed state. Thus the
first order perturbation theory is totally inadequate to this case.
In the first two cases the perturbation corrections were 10 orders of magnitude smaller
that the non-perturbed energy, so un these cases the first order perturbation theory is
applicable