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Normal modes for N = 6

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					Quantum Tunnelling:                       A B  C  D      (i)
                                                    
                                         A  B   (C  D) (ii)
                                                    ik
                                   1     i     i  
   Adding (i) and (ii) gives   A  2 1  C  1   D 
                                          k        k 

                         C exp(  a)  D exp(a)  F exp(ika) (iii)
                                                  ik
                     C exp(  a)  D exp(a )   F exp(ika ) (iv)
                                                      
                                    1       ik 
 Adding (iii) and (iv)..       C   2 1        F exp ik   a 
                                            
                                    1       ik 
 Subtracting (iii) and (iv).   D   2 1        F exp ik   a 
                                            
 Therefore:
                 i  ik              i  ik           
A  F exp( ika) 1  1   exp( a)  1  1   exp( a)
   1
   4
                   k                  k             
Transmission through barrier
 Sometimes it is convenient to use:

    sinh( x)  1 (ex  e x ) cosh( x)  1 (ex  e x )
               2                         2



Then transmission coefficient is:

     | F |2                      4 k 2 2
  T        
     | A| 2
               
              4k 2 2 cosh 2 (a)  (2  k 2 ) 2 sinh 2 (a)   
Quantum tunnelling, I: Alpha Decay
                                                    repulsive
                                        V(r)
                                                    Coulomb
Many unstable nuclei decay by                       potential
emitting alpha particles.
                                               V0

Particles in states with E>0 but E<V0    0
may tunnel through the potential                                r
barrier.                                             attractive
                                                     nuclear
                                                     potential
Quantum tunnelling, II: Nuclear Fusion
                                       V(r)
                                              Coulomb
The source of energy in the Sun
is nuclear fusion.
                                              potential
Two protons collide to form a     V0
deuteron.
Barrier is 1 MeV, but thermal           0
energies of protons is 1keV                   nuclear r
( 107K ).
                                              potential
Formal Structure of QM




                     ( eg Rae, Chapter 4)
Eigenvalue equations
Eigenvalue equations are of the form:
                 ˆ
                 On  n n

 Operator eigenfunction eigenvalue eigenfunction

Operator corresponds to a physical observable.


The eigenvalue is the outcome of a measurement of the
physical property when the system is in an eigenstate.
Operator -> eigenfunction
                                              
e.g. Momentum operator                px  i
                                      ˆ
                                              x
                                              
Eigenvalue equation:               p   i 
                                                 p
 - momentum eigenfunction.                   x
Rearrange and integrate:                 i p
                                           x
Giving:               i px
             ln( )       C
                        
Therefore momentum eigenfunction
is                                  i p x
                            A exp        A exp(i k x )
                                     
    How to tell if a function is an
    eigenfunction of some operator
      E.g. wave function                                      
                                  ( x, t )  exp i k x   t 
                                                 
   And momentum operator                 px  i
                                         ˆ
                                                 x
 Can we “extract” the momentum
 eigenvalues from the wavefunction
 with the operator?
                            
                        i   i (i k ) exp(i [k x   t ])  ( k )
                           x
Similarly we can “extract” the
energy by operating with:
                          
                        i   i (i  ) exp( i[k x   t ])  ( )
                          t
Hermitian operators

Operators correspond to physical observables,
so their eigenvalues must be real .
Hermitian operators have real eigenvalues.
            ˆ
An operator O is Hermitian if:

              f Ogdx
                             gO* f dx
                                 

 ˆ                             ˆ
 O*is the complex conjugate of O
f and g are arbitrary functions.
Hermitian operators have “orthogonal” eigenfunctions
Orthogonality
Eigenfunctions of Hermitian operators form “orthonormal”
sets.    (Orthonormal= Orthogonal + Normalised)
Orthogonality condition:

                   m  n d x   m ,n    - an overlap integral
                     *


             *m,n= 1 if m = n and *m,n= 0 if m  n
No overlap of eigenfunction nm onto the eigenfunction nn.
Necessary condition for an eigenfunction, since if a
particle is in a particular eigenstate, there must be no
possibility of finding it in a different eigenstate.

Normalisation condition (valid where m=n) :

               n n d x   |  n |2 d x  1
                 *
The Expansion Theorem



                                  {nn }is a complete set of
R is an arbitrary
wavefunction
                        cnn   orthonormal eigenfunctions
                         n        of a particular operator
Examples of expansion theorem, I:
Particle in an infinite potential                       1        (x)
well has wavefunction:
              15       x 
                           2                          0.5
   ( x)        1     
             16a       a                         u1(x)
                                                        0
What is the probability of finding
                               -1    -0.5      0                 0.5    x/a   1

the particle in the ground state? u ( x )  1 cos x 
                                                  
i.e. with eigenfunction
                                   1
                                             a    2a 

           cnun
                                                a
Write                   , then P  | c1 |  |    u1  d x | 2
                                         2         *
                                1
              n                                 a
                                        2                    2
                        x  x           2 15  16
                                  2
         15 1 a
 P 
   1
         16 a a cos  2a  1  a 2  dx   3
                                   
                                                       0.99856

  - large overlap.
Expectation value
i.e. mean value of the eigenvalue



For eigenvalues 8n of the operator O andˆ
probabilities | cn |2 for each eigenvalue, mean value of 8 is:



                        | cn |2 n
                            n
                                    
                      ˆ                    ˆ
                 O             * O d x
                                    

    Mean eigenvalue        Expectation value of operator
Proof of expectation value
 (a use of the expansion theorem)




                                                      

      ˆ
     *O d x        cm mO cn n d x 
                         * * ˆ
                                               cmcn n  m n d x
                                                 *         *

                 m          n             m ,n      

                                                      orthogonality
                                                      condition

          cmcn n  m,n   | cn |2 n    
             *

          m ,n                      n
Formal Postulates of Quantum Mechanics

 (1) A physical system consisting of one particle is
     described by a wavefunction R(x, y, z, t), from which all
     possible predictions about the physical properties of
     the system can be obtained.
    If R is normalised, the probability of finding the
    particle at position x, y, z in a volume element dxdydz at
    time t is

               | ( x , y , z, t )| d xdy d z
                               2
Formal Postulates of Quantum Mechanics

(2) The wavefunction evolves in time according to
Schrödinger’s equation:


                  2 2
           i           V ( x , y , z) 
              t    2m
Formal Postulates of Quantum Mechanics

  (3) Each physical observable (measurable quantity) has
  a corresponding Hermitian operator, whose eigenvalues
  represent the possible results of carrying out a
  measurement of the physical observable.
      Immediately after measurement, the wavefunction
  of the system will be the eigenfunction corresponding
  to the measured eigenvalue
Formal Postulates of Quantum Mechanics
(4)                     cnn
                          n



 Measurement using the Hermitian operator O results
                                          
 in the value 8n with a probability |cn|2


 The average result of a large number of identical
 measurements is

                              
                   
                  O             
                                 * O d x
                          
Heisenberg’s Uncertainty Principle
Measurement of the position of a particle leads to an
uncertainty in the particle’s momentum.
                   p x  
Such pairs of observables are said to be “incompatible”.
Another example of incompatible observables are energy
and time:
                   E t  

 The Uncertainty Principle does not apply to all pairs of
 observables, for example momentum and energy of a
 particle are “compatible observables”.
 Compatible observables
Suppose we have two observables p and q, with associated
          ˆ     ˆ
operators P and Q .



 p and q are compatible observables if the corresponding
                   
 operators P and Q share the same eigenfunctions.



For compatible observables the
commutator of the operators must vanish:     P, Q   0
                                               

              i.e.    ˆˆ ˆˆ
                      PQ  QP  0
 Generalised Uncertainty Principle
   Deviation from mean :          x  x  x 
 Uncertainty x is the RMS (root mean square) deviation:

         x     ( x) 2       x2    x  2 

Uncertainty in other observables is defined similarly.
We replace mean values by expectation values of operators
e.g. the uncertainties in the quantities p and q would be:

  p       P
              2    P  2 q 
                                                 Q2    Q  2 
                                                          


                    p.  q     1          
                                     |  [ P , Q]  |
                                 2
Time dependence
  How does a wavefunction evolve with time?
  Spatial and temporal parts of the eigenfunctions can be
  separated:

            n ( x, t )  un ( x) exp(  i En t /  )

What about general wavefunctions? Well for t = 0 we can
use the expansion theorem:

                    ( x, t  0)     cn un
                                         n

The time evolution is:
                       ( x, t )   cn un exp( iEn t / )
                                     n
Time dependence contd…
We can rewrite so that the time dependence is subsumed
into the coefficients cn :

                 ( x, t )     cn (t ) un
                                n
 where
              cn (t )  cn (0) exp( i En t /  )

 The shape of the wavefunction R changes with time,
 but the probabilities P(En) don’t change since
                P(En) = |cn (t)| 2 = |cn (0)| 2
Time dependence of expectation values
                                    
 Expectation value of an operator Q Time independent
 associated with a physical quantity q .

                                                          
       
  d  Q   d                              *                    d x
                 * Q d x 
               
                       
                                          t Q d x         *Q  t
     dt    d t                                        

 Use Schrödinger eqn.

     H    i          and             (H  )*   i   *
                 t                                           t
              d Q  1 
                                                   
                                           d x    * Q 1 H d x
 Then:                        ( H ) * Q                    
                  dt      i                              i
                               
                           i     ˆ  ( H ) *d x  i        ˆˆ
                               Q     ˆ
                                                         *QH d x
                                                     
 Since the Hamiltonian is hermitian
 we may replace     Q ( H ) * by  * H (Q )   * H Q
                                                  
                     ˆ
Time dependence of  Q 
We now find:                              
                                                              
               dQ    i
                         * H Q d x 
                                
                                                 * Q H d x 
                                                      
                dt      
                                                           
                                                              
 Finally we have the result

     d Qˆ  i                        i
                        ˆ ˆ
                *  H , Q  d x         ˆ ˆ
                                         H ,Q
                                              
       dt       
                                      
                                       

 This is an important result since it allows us to calculate
 equations of motion of operators.
Conservation Laws                   ˆ
                                d Q  i   ˆ ˆ
                                       H ,Q
                                              
                                  dt
If an operator commutes with the Hamiltonian, then its
expectation value is constant.
This means that the corresponding physical observable
is conserved.
Schrödinger’s Equation in Three
Dimensions
 For solutions to realistic problems we must generalise to
three dimensions:
      2 2
         u ( x , y , z )  V ( x , y , z ) u( x , y , z )  E u( x , y , z )
      2m

 Simple example: Particle in a box
                                                   0  x  Lx
                                                   
            V ( x , y , z)  0        if           0  y  L y
                                                   
                                                   0  z  Lz
                                                                                  Lz

            V ( x , y , z)          otherwise
                                                                                 Ly
                                                                           Lx
Separation of variables                                 0  x  Lx
                                                       
Within the box:     V ( x , y , z)  0                  0  y  Ly
                                                       
                        2                               0  z  Lz

So TISE is:                 u( x, y, z )  E u ( x, y, z )
                              2
                     2m
 Write:       u( x, y, z)  X ( x)Y ( y) Z ( z)
          2
             d2X        d 2Y       d 2Z 
           YZ   2
                     ZX     2
                                XY    2
                                                        E XYZ
         2m  d x        dy         dz 

 Divide through by XYZ :

                   2  1 d 2 X 1 d 2Y 1 d 2 Z 
                            2
                                      2
                                              2 
                                                   E
                  2m  X dx       Y dy     Z dz 
 Separation of variables, contd…
                  2  1 d 2 X 1 d 2Y 1 d 2 Z 
                           2
                                     2
                                             2 
                                                  E
                 2m  X dx       Y dy     Z dz 

As before, equation is true over all space. So each term must
be a constant, the sum of which equals the energy E .

        1 d2X                       1 d 2Y                        1 d 2Z
                k x2                      k y
                                                2
                                                                          k z2
        X dx 2                      Y dy 2                        Z dz 2


                                                 
                              2
 with                              k x  k y  k z2  E
                                     2     2
                            2m
 General solution is of type
                         X ( x)  A sin( k x x)  B cos( k x x)      etc..
Boundary conditions:
Apply BCs to each dimension, as in 1D infinite well and find..

               nx         n y        nz
          kx       ; ky       ; kz 
                Lx          Ly          Lz

  With the n’s being integers

 Eigenfunction is then:
                                       n x  x   n y  y   nz  z 
                 u( x , y , z)  A sin               L  sin L 
                                                 sin
                                       Lx   y   z     
and eigenvalue is

                                                          nx n 2 nz 
                                                
                         2                             2   2 2       2
                  E          k x  k y  k z2
                                2     2
                                                          2   2  2
                                                                 y
                       2m                              2m  Lx Ly Lz 
                                                                      
Degeneracy
                                                                        3   18
        
                                          
               2   2
                                                                        3   17
    E           n x  n y  n z2
                   2     2
                                                       V  L x L y Lz
       2mV 2 / 3                                                        6   14
A given energy level may have several                                   1   12
eigenfunctions. This is called degeneracy.
                                                                        3   11
                                                                        3   9
    {nx, ny, nz}       ( nx2+ ny2+ nz2 )       Degeneracy
                                                                        3   6
       1,1,1                  3                    1
       2,1,1                  6                    3                    1   3
       2,2,1                  9                    3
       3,1,1                  11                   3
       2,2,2                  12                   1         Degeneracy Energy
       3,2,1                  14                   6
                                                                        in units
                                                                        of 2 2
       3,2,2                  17                   3
       4,1,1                  18                   3
       3,3,1                  19                   3
                                                                                 2mV 2 / 3
Particles in a Central Potential
Central potential depends only on the distance of the
particle from the origin, V(r).

Examples:
Harmonic oscillator             V (r )    1
                                           2   K r2
                                                 e2
Electron in a hydrogen atom     V (r )  
                                               4 0 r

 The central potential problem is the foundation for
 understanding all of atomic and nuclear structure, and
 spectroscopy.
Transforming to Spherical Polar
Coordinates
  z
              Transforming from x,y,z, to r,,
                        x  r sin  cos 
                        y  r sin  sin 
                       z  r cos
                 y
      


          x
Schrödinger’s Equation in Spherical
Polar Coordinates
The time independent Schrödinger equation becomes:
   2
      1   2 u      1            u    1     2u 
     2     r     2         sin     2 2       2
                                                           V (r )u  Eu
  2  r  r   r  r sin          r sin    


Mass of particle

 The general solution has the form:
                 u (r , , )  R(r )Y ( , )
 Radial wavefunction:                   Angular wavefunction:
 determines the energy levels           this determines the
 (in absence of magnetic field)         angular momentum states
Angular momentum                                 L
                                        p
 Classical definition: L  r  p                          r

 For a central potential the
 angular momentum is conserved.

Angular momentum operators
 Lx
                                 
   i   sin    cot  cos         i  
                                        Lz
                                          
                                   
   i    cos   cot  sin   
 Ly
                               

The operator for the z – component of the angular
momentum is the simplest. Therefore we use the z – axis
as our “quantisation axis” whenever possible.
Total angular momentum
One further important operator, namely the square of the
angular momentum:
                            
                L2  L2  L2  L2
                       x    y    z

conventionally called the “total angular momentum” operator.
In spherical polar coordinates it becomes:
                     1                   1 2 
         
         L2    2            sin            2
                     sin           sin   
                                               2



It has the same form as part of the Laplacian operator:

      1   2       1                 1     2
 2  2    r     2         sin     2 2
     r  r   r  r sin          r sin    2
Another way of writing Laplacian




          1   2    1 2
        2
        2
               r    2 2 L
         r  r   r  r
Separation of variables in polar
coordinates
Time independent Schrödinger equation :
             2         2  u   1 2 
                        r      2 L u  V (r ) u  E u
            2 r 2    r   r        
Write wavefunction as a product of radial
part and angular part:    u(r , ,  )  R(r ) Y ( ,  )
      2         2  R     R 2 
                 r     Y  2 L Y   V (r ) RY  E RY
     2 r 2    r   r           
 Divide both sides by u = RY :
          2  1   2  R     1 1 2 
            2       r      2 L Y   V (r )  E  0
         2 r  R  r   r   Y      
                     only depends on only depends on the
                     the r coordinate angle variables 2 and N
Separation of variables…
Eigenvalue equation for total angular momentum:
                    
                    L2 Y ( ,  )   2  Y ( ,  )

                     choose     as the eigenvalue
                                      2

Use this result to eliminate the angular dependence in TISE
           2       1   2  R   1 1 2 
                       r       2 L Y   V (r )  E  0
          2 r 2   R r r  Y          
                                            
                              Angular part appears as an effective
Multiply through by R:        “centrifugal” potential


        2   2  R( r )     2            
              r               2  V ( r )  R( r )  E R(r )
      2 r  r 
          2
                    r   2 r                
                   Radial Schrödinger equation (see H-atom……)
Eigenstates of Angular Momentum
 Operator for z component of angular momentum:
                i  
              Lz
                          
                                          Y   i   Y   mY
                                         Lz
 Eigenvalue equation for Lz :
                                                    
                                                      eigenvalue
 To separate the variables further we write:

         Y( , )   ( )  ( )

Eigenvalue equation becomes:             i      m  
                                    Lz
                                                     
 1 cancells on both sides:

                      i      m
                  Lz
                                    
               ˆ
Eigenstates of Lz
                                 
Eigenvalue equation is:       i     m
                                 
which has a solution:
                              ( )  A exp(i m )
A wavefunction should be single valued; which implies

                     (  2 )   ( )

Therefore:        m  0,  1,  2,  3
Eigenstates of Total Angular Momentum
Eigenvalue equation for total angular momentum:
                   
                   L2 Y ( ,  )   2  Y ( ,  )
This is shorthand for:
                1              Y         1  2Y 
   
  L2 Y    2            sin                       Y
                                                           2

                sin            sin 2   2 
Substitute                   Y( , )   ( )  ( )
                                               2Y     2
 Using form of Lz eigenfunction                      2   m 2 
                                               2    
This substitution allows us to divide
through by , leaving:
  Associated               1                      m2 
                                    sin                0
  Legendre Equation      sin                  sin  
                                                         2
Solutions of Legendre Equation:
Case of m = 0 is called the Legendre equation:

                1            
                        sin        0
              sin          
Solutions are polynomials in even or odd powers of cos().

                      P  cos( ) 

                            R is order of
            polynomial determined by the        (  1)
            condition
So the eigenvalue of total
angular momentum is:
                               2
                                  ( 2
                                               1)
Solutions for m  0

             1                      m2 
                      sin                0
           sin                  sin  
                                           2



Solutions are given by:

                                           d |m| P
                      
           P|m|  x   1  x    
                                2 |m|/ 2
                                           d x|m|
where
        x  cos( )
            ˆ 2 operator:
Summary for L
   Eigenvalue equation:      ˆ
                             L2 Y ( , )     2
                                                    (  1) Y ( , )

   Where      Y( , )   ( )  ( )        and      is an integer.
                                                       1
                 P |m|
                          cos( )       ( ) 
                                                       2
                                                          exp(i m )

 So the eigenfunctions of total angular momentum are:


        Y m  ,   A m P (cos )exp(im )
                               |m|




            These functions are called the
            Spherical Harmonics
Spherical Harmonics                      Y m  ,   A m Pm (cos )exp(im )



               Y00  (4 )1/ 2

                                              Y11  (3/8 )1/ 2 sin exp(i  )

        Y10  (3/ 4 )1/ 2 cos


                                           Y22  (15/ 32 )1/ 2 sin 2  exp( i 2 )

Y20  (5/16 )1/ 2 (3cos2   1)

Dark shading: positive
                                   Y21  (15/8 )1/ 2 cos sin exp  i  
Light shading: negative
Quantum numbers for angular
momentum, R and m
                                ˆ
          ˆ 2  cannot exceed  L 2 
  Since  Lz
  then   m2  (  1) i.e.   m 


  So there are (2R +1) values of m for each R

  i.e. the degeneracy is (2R +1)
Summary of Angular Momentum
Eigenvalue Equations
Total angular momentum:

     ˆ2 Y ( , ) 
     L ,m                 2
                              (  1) Y ,m ( , )
 where R is a positive integer = 0, 1, 2, 3…

z - component of the angular momentum

          ˆ
          LzY ,m ( , )  mY ,m ( , )
 where m is an integer = 0, 1, 2, …., with m in the
 range   m 
Semiclassical Model for
Angular Momentum
Visualise orbital angular momentum L as a vector of length
 (  1) whose projection along the quantisation axis is mħ

                                        2

        z
                   1                    1


                   0                    0


                  1                   1



               Radius = 2ħ            2

                                               Radius = 6ħ
                   l=1 1               2
                                       l=2
 The Hydrogen Atom             Coulomb              e 2
                               potential
                                          V (r ) 
                                                   4 0 r
                         2 2      e2
 Schrödinger equation:      u         u  Eu
                         2      4 0 r
    Reduced mass:
     me m p /(me  m p )                 u(r, , )  A R(r ) Y ,m ( , )
 Radial Schrödinger equation is:
        2
           2  R(r )   (  1)                        2
                                                                  e2
           r                           R( r )  E R( r )
   2 r  r 
       2
                  r   2 r 2      4 0r 
                                   8 E           2 e 2
With substitutions:    r ;    2 ;  
                               2
                                               4 0 2
(E is negative for bound states as potential is zero for infinite r)


radial eqn.              1   2 R    1 (  1) 
becomes:                              4     R 0
                           
                          2
                                                  
                                                    2
Solution of radial equation
   Solution is of the form:          R(  )  F (  ) exp(  / 2)
                                                             k
   where F() is a polynomial:                 F ()   ap  p
                                                         p 0
with:     k 1  n         where n = 1, 2 ,3 …principal quantum num
(Highest term is when k (=   R ) =n -1 )

                             8 E                 2 e2
 As    r ;                     ;                             n
                     2
                                 2
                                                4 0   2


 Then energy eigenvalues for the hydrogen atom are:
                                                                 2
                            2
                                 2 e  1
                                 2         2         2
                   E               2
                       8    8  4 0  n 2
Energy Eigenvalues:
                                                  n=
                                 2                n=4
                 2 e  1
                   2         2
                                  E1              n=3
         En               2  2
               8  40 2  n
                                n               n=2

N.B. Energy (in absence of magnetic field) only
depends on the principal quantum number n, not
on the values of R and m.
This means that the energy levels are
degenerate.

The value of E1 = 13.6 eV.
This is the ionisation energy for hydrogen.       n=1
Quantum numbers for hydrogen contd…
Principle quantum number n= 1,2,3,4,5…
Determines energy levels and radial extent of wavefunction.

Angular momentum quantum number R is integer, but R 
n –1
Determines value of total angular momentum

Magnetic quantum number m can have 2R +1 values between

-R and + R.
Determines component of angular momentum.

Example:    n= 1, R = 0                             n1

            n= 2, R = 0 , 1
                                          F ()     ap  p
                                                    p
Principal quantum number determines
radial extent of wavefunction
We plot here the probability density Pn,l(r)dr of finding the
electron in a spherical shell of radius r and thickness dr:
                             Pn, (r )  4 r | An, | | Rn, (r ) |
                                                     2           2             2

                   0.6
                                                     In excited states
                              n =1, l= 0             electron moves further
                   0.4
                                                     away from nucleus
    Pn,l (r/a0 )




                                        n =2, l= 0
                   0.2                                           n =3, l= 0



                    0
                         0          5                    10               15       20
                                                         r/a 0
Effect of increasing the angular
momentum quantum number R
                 0.15
                                        n=3, l=1

                            n=3, l=2                    n=3, l=0
                  0.1
    Pn,l(r/a0)




                 0.05



                   0
                        0         5     10         15          20
                                       r/a 0
Spectroscopic Notation
By convention the orbital angular momentum R in
atomic states is labelled by a letter, rather than a
number:

 Orbital quantum number:    R = 0,    1,    2,     3,     4
 Spectroscopic notation:       s      p     d      f      g


 The notation goes back to the early days of atomic
 spectroscopy, when different spectral series (“sharp”,
 “principal”, “diffuse”…) were associated with atomic
 states with different angular momenta
Degeneracies of energy levels in
the hydrogen atom
      n=           1       2          3              4

    States        1s    2s, 2p    3s, 3p, 3d   4s, 4p, 4d, 4f

    Orbital        1     1+3      1+3+5        1+3+5+7
  degeneracy              =4        =9           =16
   Including
Spin degeneracy    2       8         18             32

 “Spin degeneracy” relates to an extra intrinsic degree of
 freedom for the electron – more of this later!

 Note that the degeneracy of the n’th level = 2n2
Spectroscopy of hydrogen
                                        E1
Energy levels of hydrogen are     En   2
                                        n
                e4 
where
               2(4 )2   13.6eV .
         E1             
                    0    
In transition from n=p to n=q, emitted
photon frequency is:


     E1  1 1      1   1 
     2  2   R 2  2 
      q
           p 
              
                   q
                       p 

R is called Rydberg’s constant,
and has the value
3.2880461015 Hz.

                                             Balmer series: q =2

				
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