HYDROGEN ATOM VISIBLE SPECTRUM by klutzfu51

VIEWS: 133 PAGES: 39

									Fig. 4-9, p. 125
 HYDROGEN ATOM VISIBLE SPECTRUM




This grouping of lines has a name. It is called the Balmer Series.
(1885, Johann Balmer )
Notice how the spacing increases as the wavelength gets longer
This will become important later.
•     the energy levels are found to follow the
     empirical formulas given by Balmer and
     Rydberg.
                                      ⎛ Z 2 ⎞
          E = − 2 . 178 x 10 − 18 J ⎜ 2 ⎟
                                      ⎝ n ⎠
Z = atomic number (1 for H)     n = integer (2, 3, ….., ….)
    • R = -2.178 x 10-18 J is the Rydberg Constant
• We can use the equation to predict what ΔE is
  for any two energy levels
                ΔE = E final − E initial
                  ⎛ 1 ⎞               ⎛ 1 ⎞
              −18                 −18
ΔE = −2.178x10 J⎜ 2 ⎟ − (−2.178x10 J)⎜ 2 ⎟
                  ⎜n ⎟
                  ⎝ final ⎠           ⎝ n initial ⎠

                             ⎛ 1        1 ⎞
          ΔE = −2.178x10−18 J⎜ 2 − 2 ⎟
                             ⎜n
                             ⎝ final n initial ⎟
                                               ⎠
           • H Atom Energy level diagram Visible Spectrum
  Energy
   0                                       Note: As n goes to
                                 n=∞
-R/25                            5         Infinity energy levels
-R/16                            4          get closer and closer
 -R/9                            3         together.Looks more
                                           Continuous
 -R/4                            2
              Balmer                                 R
                                          En i = −        2
                                                     ni
                                                      R
                                          En f = −        2
                                                     nf
                                                              ⎛ 1  1 ⎞
                                     ΔE = E n f   − En i = − R 2 − 2 ⎟
                                                              ⎜
                                                              ⎜n   ni ⎟
                                                              ⎝  f    ⎠

  -R                             1
Fig. 4-13, p. 129
                  ⎛ Z2 ⎞
E = −2.178x10−18 J⎜ 2 ⎟
                  ⎝n ⎠

  • Energy levels get closer together
    as n increases


  • at n = infinity, E = 0
         Bohr Model of the Atom
• Postulates:
1. The coulomb force on a planetary electron provides
   the centripetal acceleration required for a
   dynamically stable circular orbit.

2. The only permissible orbits are those in the discrete
   set for which the angular momentum
   of the electron equals an integer
    n times h, where h = h / 2π
                 Note Quantization of Angular Momentum
          Postulates Part II
3. An electron moving in one of these
   stable orbits does not radiate.
4. Emission or absorption of radiation
   occurs only when an electron makes a
   transition from one orbit to another.
       The Bohr Hydrogen Atom
• We can use Bohr’s postulates to derive the
  energy levels in Hydrogen
• Postulate 1 implies:
                               Z=1

                      2      2
                   mv  ke
                      = 2
                    r  r

   From the second postulate we can write
                    mvr = nh
 Radius of the Hydrogen Atom
• We get the equation
                   2 2
                  n h     2
             r=       2
                        = n a0
                  mke
 Which if evaluated comes out as:

 a0 = 5.29 10-11 m ≈ 0.53Å
Energy Levels in the Bohr Atom
• The kinetic energy of an electron in circular motion is:
                                   2        2
              1 2    ke   ke
           K = mv = r 2 =
              2      2r    2r
  The potential energy in a Coulomb Potential
                               2
                         ke
                    V =−
                          r
   The total energy is then
                               2        2          2
                     ke   ke     ke
        E = K +V = −    +     =−
                      r    2r     2r
         Energy Levels Part II
• The energy of a given level depends on
  the radius of the electron:   Actually “Reduced Mass
                                   m m /(m + m )
                                         e p   e   p
                     2          2 4
                ke            mk e      w0
        En =      2
                         =−     2 2
                                    =− 2
               2n a0          2h n    n
                       2.178 eV
                 w0 = 13.60 x 10 -18 Joules
Fig. 4-13, p. 129
• Example: At what wavelength will emission
from
    n = 4 to n = 1 for the H atom be observed?
                           ⎛ 1        1 ⎞
        ΔE = −2.178x10−18 J⎜ 2 − 2 ⎟
                           ⎜n
                           ⎝ final n initial ⎟
                                             ⎠
                                         1        4
                                −18    ⎛  1⎞
     ΔE = −2.178x10                   J⎜1− ⎟ = −2.04 x10−18 J
                                       ⎝ 16 ⎠

                           hc
ΔE = 2.04 x10   −18
                      J=                     λ = 9.74 x10−8 m = 97.4nm
                           λ
             Alternative Units
• 1 electron volt = 1.602 x 10-19 joules

• 2.18 x 10 – 18 joules/ 1.602 x 10-19 joules/eV

               = 13.6 eV
  • We can use the equation to predict the
      Photon Energy Connecting
   any two energy levels


                  ΔE = E final − E initial
                  ⎛ 1 ⎞               ⎛ 1 ⎞
              −18                 −18
ΔE = −2.178x10 J⎜ 2 ⎟ − (−2.178x10 J)⎜ 2 ⎟
                  ⎜n ⎟
                  ⎝ final ⎠           ⎝ n initial ⎠

                               ⎛ 1          1 ⎞
Ephoton =   ΔE = −2.178x10−18 J⎜ 2 − 2 ⎟
                               ⎜n
                               ⎝ final   n initial ⎟
                                                   ⎠
                               ⎛ 1     1 ⎞ eV
                Ephoton = 13.6⎜ 2 − 2 ⎟
                               ⎝n m ⎠
       HYDROGEN ATOM
          IONIZATION
            Energy

           = R/n2 = 2.179874113 x 10
                                     -18 J
0.00
Fig. 4-12, p. 129
       Extension to Higher Z
• The Bohr model can be extended to any
single electron system….must keep track of Z
      (atomic number).

                    ⎛ Z2 ⎞   Z = atomic number
  E = −2.178x10−18 J⎜ 2 ⎟
                    ⎝n ⎠     n = integer (1, 2, ….)


• Examples: He+ (Z = 2), Li+2 (Z = 3), etc.
 Extension to Higher Z (cont.)

• Example: At what wavelength will emission
from
    n = 4 to n = 1 for the He+ atom be observed?
                                ⎛ 1        1 ⎞
      ΔE = −2.178x10−18 J (Z 2 ) 2 − 2 ⎟
                                ⎜
                                ⎜n
                                ⎝ final n initial ⎟
                                                  ⎠
                              2    1      4
                                  ⎛  1⎞
    ΔE = −2.178x10     −18
                             J (4)⎜1− ⎟ = −8.16x10−18 J
                                  ⎝ 16 ⎠
                       hc
            −18
 ΔE = 8.16x10     J=                λ = 2.43x10−8 m = 24.3nm
                       λ
                                           λ H > λ He   +
 Albert Einstein explained the
photoelectric effect

Explanation involved
light having particle-like
behavior.
   –
Fig. 4-14, p. 132
     EM Radiation Has Wave
            Properties


                  λν = c
EM Waves Have Particle Properties

                                     hc
                      E = hν or E=
                                      λ
                               =
            h = Planck’ constant 6.626x 10-34 J ⋅ s
                      s
                     Momentum P = h/λ
Next Advance in Understanding Atomic Structure Came
From Asking “If EM Waves Have Particle Properties-Do
      Other particles than the photon Have Wave
                      Properties” ?

 • yes- Particles also exhibit duality
De Broglie Associated a Wave Length
         With Particles


  Planck’s Constant   Mass of particle




   λ = h/(mV)
                      Velocity of particle
Wavelength of an electron traveling 1.24 x 107 meters/sec
                                    0.041 x speed of light
λ = h/(mV)
        =         (6.626 x 10-34 Joule sec )
        (9.11 x 10-31 kilograms x 1.24 x 107 meter/sec)
                 = 5.87 x 10-11 meters
      comparable to interatomic distances

     Wavelength of a baseball traveling 92.5 miles/hour
                                1.4 x 10-7 speed of light
λ = h/(mV) =       (6.626 x 10-34 Joule sec )
             (0.149 kilograms x 41.3 meter/sec)
                  = 1.08 x 10-34 meters
  Much smaller than baseball or distance to home plate
Bohr’s Model was improved upon
   in the 1920’s
“WAVE MECHANICAL” or
“QUANTUM MECHANICAL” Model.




                   An “Operator Equation” – performing the operation
                    indicated by H on the wave function leads to a simple
                   answer , the energy times the wave function


                       They govern the motion of the electrons

       IN PARTICULAR

                                    IF CORRECT THEN OUR JOB
                                     IS TO DEFINE H
                                    AND THE WAVE FUNCTION
                                    H Must Contain the Interaction terms
                of the electrons
•   Schrödinger proposed a Wave Equation to describe how
    electrons move in atoms
•   It has a form   HΨ=ΕΨ
•   Where H is the Hamiltonian “Operator” *
    Ψ is the Wave Function       E is the energy of the system




*Simple Example of Operator equation:
In the equation 2a = b, the “operation” is multiply by 2.
   Operating on a gives b.

     Operations can be more complicated
STANDING WAVES




                 Fig. 4-17, p. 135
STANDING WAVES




                 Fig. 4-18, p. 135
Particle in a One Dimensional Box




                             Fig. 4-24, p. 146
H




    GENERAL SOLUTION OF THIS DIFFERENTIAL EQUATION
    General
    Solution




    AT ENDS OF BOX     DEMAND Ψ = 0 WHY?



               TRUE WHEN   INTEGER

                                     OR



0   L

        SO
If Wave Function squared is probability then integrated over all space must equal 1




                                   SO




             We Can Now Plot These Solutions For n =, 1.2, 3 etc.
1



2




3




    Fig. 4-24, p. 146
Fig. 4-24, p. 146

								
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