1. What’s coming up??? Nov 3,5 Postulates of QM, p-in-a-box Ch. 9
Oct 25 The atmosphere, part 1 Ch. 8
Oct 27 Midterm … No lecture
Oct 29 The atmosphere, part 2 Ch. 8
Nov 1 Light, blackbodies, Bohr Ch. 9
Nov 3,5 Postulates of QM, p-in-a-box Ch. 9
Nov 8,10,12 Hydrogen atom Ch. 9
Nov 15 Multi-electron atoms Ch.10
Nov 17 Periodic properties Ch. 10
Nov 19 Periodic properties Ch. 10
Nov 22 Valence-bond; Lewis structures Ch. 11
Nov 24 Hybrid orbitals; VSEPR Ch. 11, 12
Nov 26 VSEPR Ch. 12
Nov 29 MO theory Ch. 12
Dec 1 MO theory Ch. 12
Dec 2 Review for exam
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2. The wave is “quantized”. It has nodes.
A 1-D standing wave.
The wave is “quantized”.
It has nodes.
The equation that determines
Now use this with the de Broglie relationship
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3. ( )2 Y d 4p2p2 = - Y dx PARTICLE IN A BOX h2 d dx h p Y = - 2 2 NOW...
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4. ( )2 PARTICLE IN A BOX d dx h p Y = - The particle has KE momentum
Now write equation to give KE..
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5. ( )2 PARTICLE IN A BOX d dx h p Y = - Rewriting 2
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6. PARTICLE IN A BOX Rewriting…. This equation determines the KE
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7. PARTICLE IN A BOX Including the potential V(x)…... Etotal = EK + V(x)
No confine the particle to a BOX……...
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8. PARTICLE IN A BOX POTENTIAL ENERGY OUTSIDE BOX V = w INSIDE BOX V = 0
We have…….
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9. PARTICLE IN A BOX The SCHRODINGER WAVE EQUATION for a
for a 1-D particle with no potential acting on it!
Rewriting
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10. PARTICLE IN A BOX Y = A kx sin A possible solution is…….
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11. Y (x) A kx = sin PARTICLE IN A BOX ENERGY Y(0) = 0 Y(L) = 0 x Lx L
BOUNDARY CONDITION
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12. PARTICLE IN A BOX WE KNOW Y(0) = 0 x = 0 BOUNDARY CONDITIONS x = L and
Y(L) = 0
THEN…..
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13. = Y A kx sin kL n = p PARTICLE IN A BOX Y (L) = A sin kL = and
x = L
WE KNOW Y
(L)
THEN….. = A
sin kL = kL n = p
TRUE WHEN
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14. Y (x) A L x = sin p p n = k L kL n = p PARTICLE IN A BOX n
Why can’t n be zero????????
The wave function would be zero so, no particle!!!!
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15. What does Y Mean????? The answer lies in WAVE-PARTICLE DUALITY
Electrons have both wavelike and particle like properties.
Because of the wavelike character of electron
we CANNOT say that an electron
WILL be found at certain point in an atom!
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16. WAVE-PARTICLE DUALITY
For small particles such as electrons
if we accurately know their position,
we know almost nothing about their momentum,
and therefore their velocity
and vice versa.
You cannot pin an electron down!
Heisenberg understood this!
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17. THE HEISENBERG UNCERTAINTY PRINCIPLE He postulated that if...
Dx is the uncertainty in the particle’s position
Dp is the uncertainty in the particle’s momentum
For a particle like an electron, we cannot know
both the position and velocity
to any meaningful precision simultaneously.
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18. Suppose we can estimate position to 1% of its radius
BASEBALL m= 0.2kg
Suppose we can estimate position to 1% of its radius
Typical baseball radius is 0.05m
= 5 x 10-31m/s
VERY SMALL UNCERTAINTY!!!!!!
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19. Suppose we can estimate position to 1% of radius of H-atom
ELECTRON
me = 9.11x10-31kg
Suppose we can estimate position to 1% of radius of H-atom
1 nm = 10-9 m
r = 0.05 nm
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20. This means we have no idea of the velocity of an electron if we try to tie it down!
Alternatively if we pin down velocity we have no idea where the electron is!
So for electrons we cannot know precisely where they are!
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21. The answer is in the interpretation of those wave functions from……….
So what do we do??????
The answer is in the interpretation of those wave functions from……….
THE SCHRODINGER EQUATION…..
The wavefunctions () are merely mathematical expressions!
Like this!
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22. Y = A kx sin FOR A PARTICLE IN A BOX
OR AN ELECTRON IN THE GROUND STATE OF A H-ATOM
Max Born told us how to interpret these wavefunctions () ……….
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23. Max Born told us how to interpret these wavefunctions () !
we do not no precisely where the electron is
we speak of the likelihood of where the electron is!
a probability that the electron is at certain point in space!
What is that probability?????
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24. THE BORN INTERPRETATION!
The probability of finding an electron at a given location is proportional to the square of .
If  = 0.1 at one location then…..
The probability is 0.01.
If = at another location then…..
The probability is 0.04.
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25. Y (x) A L x = sin p PARTICLE IN A BOX n
The probability of finding the particle on a segment of the x-axis of length dx surrounding the point x is….
If we sum all of these infinitesimal probabilities, the
total must be equal to one, since there must be
some finite probability of finding the particle somewhere..
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26. We do the sum by integrating the square of the
wavefunction from x=0 to x=w, and setting the result
equal to unity Y2
(x)dx w w A2 n
x dx
sin2 p = n L
= 1
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27. Y (x) L x = 2 sin p n When we carry out the integration over Y2 we
can then solve for the unknown A Y n
(x) L x = 2
sin p
What do the wavefunctions look like?????
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28. 2 n p Y (x) = sin x L L ENERGY n = 3 What are the ENERGIES??? n = 2
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29. ENERGY OF A PARTICLE IN A BOXL
Each wave function is..
n = 3
An integral number of
n = 2
HALF-WAVELENGTHS
n =1
SO….
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30. ENERGY OF A PARTICLE IN A BOX
INTEGRAL NUMBER OF L
HALF-WAVELENGTHS
n = 3
n = 2
So what is the energy???
n =1
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31. ENERGY OF A PARTICLE IN A BOX
KINETIC ENERGY OF THE PARTICLE IS
USING THE de BROGLIE RELATIONSHIP
GIVES….
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32. ENERGY OF A PARTICLE IN A BOX
KINETIC ENERGY OF THE PARTICLE IS
NOW USE THE WAVELENGHTH
GIVES…...
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33. ENERGY OF A PARTICLE IN A BOX
KINETIC ENERGY OF THE PARTICLE IS h 2 h 2 = = E KE 2 m l 2 2 2 m (2 L / n)
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34. ENERGY OF A PARTICLE IN A BOX
KINETIC ENERGY OF THE PARTICLE IS h 2 h 2 = = E KE 2 m l 2 2 2 m (2 L / n) En
EKE
NOW CALL
WE GET
ENERGY QUANTIZED!!!!!!!!!
ENERGY LEVELS…...
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35. PARTICLE IN A BOX ENERGY n = 3 n = 2
ENERGIES QUANTIZED!!!!!!!!!!!!!!!!!!!
n =1
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36. PARTICLE IN A BOX The application of the BOUNDARY CONDITIONS
Gives a series of QUANTIZED ENERGY LEVELS
ONLY CERTAIN ENERGIES ALLOWED!
DETERMINED BY THE NUMBER n
n is a QUANTUM NUMBER
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37. THE NUMBER OF NODES INCREASES AS THE ENERGY INCREASES
THE WAVEFUNCTIONS 2 n p Y =
(x)
sin x n L L
A NODE
ENERGY
n = 3 Y
CHANGES SIGN
n = 2
n =1
THE NUMBER OF NODES INCREASES AS THE ENERGY INCREASES
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38. Y2 PROBABILITIES ENERGY n = 3 ZERO n = 2 NO CHANCE OF FINDING ELECTRON
AND….
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39. PARTICLE IN A BOX FOR ONE DIMENSION SCHRODINGER EQUATION 1-D REQUIRES
ONE QUANTUM NUMBER!
THREE DIMENSIONS
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40. ( ) { } THREE DIMENSIONS E h m n L = + 8 h d d d - + + Y ( x , y , z )
SCHRODINGER EQUATION { } h 2 d 2 d 2 d 2 - + + Y ( x , y , z ) = E Y ( x , y , z ) p 2 2 2 2 n n n n n n n 8 m dx dy dz x y z x y z E h m n L x y z = + 2 8 ( )
3-D REQUIRES
THREE QUANTUM NUMBERS!!!
NOW ON TO THE H-ATOM……..
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