13.1 Quantum & Nuclear Physics 2013


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  • Light and its nature have caused a lot of ink to flow during these last decades. Its dual behavior is partly explained by (1)Double-slit experiment of Thomas Young - who represents the photon’s motion as a wave - and also by (2)the Photoelectric effect in which the photon is considered as a particle. A Revolution: SALEH THEORY solves this ambiguity and this difficulty presenting a three-dimensional trajectory for the photon's motion and a new formula to calculate its energy. More information on https://youtu.be/mLtpARXuMbM
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  • Great summary of difficult topics. I also would love to download the slide show for my students.
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13.1 Quantum & Nuclear Physics 2013

  1. 1. Quantum PhysicsAHL 13.1
  2. 2. THE PHOTOELECTRIC EFFECT Early last century, several people had noticed that light was capable of ejecting electrons from various metal surfaces. This effect, known as the photoelectric effect, is used in photographer‟s light meters, sound tracks of motion pictures and electric eyes used to automatically open doors. Photoelectric Effect
  3. 3. THE PHOTOELECTRIC EFFECT Remember that metals have “free” electrons that are not tightly bound. It is these electrons that allow current and heat to flow in a metal, as the electron‟s move.
  4. 4. THE PHOTOELECTRIC EFFECTExamples of the freeing of electrons from light energy…  Hertz, in 1887, noticed that a spark would jump between electrodes if exposed to UV light (electrons being released from the electrode).  Other scientists noticed that the leaves of a negatively charged electroscope diverged less over time while a positively charged electroscope did not.
  5. 5. THE PHOTOELECTRIC EFFECTThis was due to the electrons “leaking” away from the electroscope as the light energy struck the top plate, allowing the electrons to escape.
  6. 6. CLASSICAL THEORYWe will now look at the predictions made for light using classical (light behaving as a wave) physics and the actual observations that were made in experiments.
  7. 7. CLASSICAL THEORYCLASSICAL ACTUAL PREDICTION: OBSERVATION:The more intense Intensity (brightness) did (brighter) the light, the not lead to high greater the kinetic velocity electrons. energy of ejection of the Instead it led to a electron. Bright light greater numbers of would eject electrons at electrons being ejected high speed from the metal.
  8. 8. CLASSICAL THEORYCLASSICAL PREDICTION: ACTUALMore photoelectrons should be ejected by OBSERVATION: low frequency radiation (i.e. red) than by high frequency radiation. Experiments showed that highe-m waves are oscillating electric and frequency (UV) radiation ejected photoelectrons more magnetic fields. readily than low frequency.Low frequency waves allow more time for There was a minimum the electron to move in one direction frequency below which no before the field reverses and the photoelectrons were electron moves in the opposite direction. ejected. This was called theHigh frequency waves would move so fast threshold frequency which is the electron would hardly begin to move different for different materials. in one direction before it was forced to reverse direction - not ideal for ejection.
  9. 9. CLASSICAL THEORYCLASSICAL ACTUAL OBSERVATION: PREDICTION: Photoelectrons wereThere should be a time ejected instantaneously. delay between when a radiation is incident on a surface and when the photoelectrons are ejected (See point 2)
  10. 10. CLASSICAL THEORYCLASSICAL ACTUAL PREDICTION: OBSERVATION:The radiation‟s wavefront By limiting the amount falls over the whole of light on a surface, surface, billions of a single electron photoelectrons should could be ejected. be simultaneously ejected.
  11. 11. CLASSICAL THEORYCLASSICAL ACTUAL PREDICTION: OBSERVATION:One velocity of ejection Emitted photoelectrons should be possible for have a range of radiation of one ejection velocities frequency. and energies.
  12. 12. CLASSICAL THEORYThe observations made on the previous slides do NOT AGREE with the predictions made by the Classical Theory. Photoelectric Effect - Changing VariablesHow can we resolve this?
  13. 13. EINSTEIN‟S EXPLANATION -PHOTOELECTRIC EFFECTIn 1905 Einstein adopted quantum theory to explain the photoelectric effect and was awarded a Nobel Prize for Physics in 1921.Planck also used quantum theory to explain the photoelectric effect. He said that the quantum effect occurred at the point where the radiation struck the electrons. The electron would only accept a discrete amount of energy from the incident radiation.
  14. 14. EINSTEIN‟S EXPLANATION - PHOTOELECTRIC EFFECT Too little energy and the electron would accept none. Too much energy, the difference would be emitted as radiation.
  15. 15. EINSTEIN‟S EXPLANATION -PHOTOELECTRIC EFFECTEinstein also said that not only was the energy absorbed and emitted by atoms in bursts but the incoming radiation was in the form of discrete entities and not a continuous wave.He named these discrete entities light quanta. This was renamed photons (a quantum of radiant energy) later as they do behave like particles and in keeping with other particles, electrons, protons and neutrons.
  16. 16. EINSTEIN‟S EXPLANATION -PHOTOECLECTRIC EFFECTEinstein‟s explanation depends on the relationship E = hfA UV photon would have more energy than a blue light photon as it has a higher frequency.The key to his explanation is that each photon on striking an atom and being absorbed may release only one electron. It never shares its energy amongst electrons.Any excess energy will be given as kinetic energy to the electron.
  17. 17. EINSTEIN‟S EXPLANATION -PHOTOECLECTRIC EFFECTAccording to Einstein, only high frequency light would have enough energy( E = hf ) to eject an electron from a metal surface.Low frequency light (like red light) might not have enough energy to pull the electron away from the atom‟s nucleus.
  18. 18. EINSTEIN‟S EXPLANATION - PHOTOECLECTRIC EFFECTImagine that photon electrons in an E = hf atom are at the bottom of a E=0 potential energy energy of electrons in an atom well that has a W hf o (work function) sloping base as -ve - shown: - -
  19. 19. EINSTEIN‟S EXPLANATION - PHOTOECLECTRIC EFFECTAs each electron is at a photon different depth, they are bound to the E = hf atom by a different amount. Each electron will then be E=0 emitted with different energy of electrons energies. If an in an atom electron absorbs a W hf o (work function) photon with sufficient -ve energy, the electron - - can be freed. -
  20. 20. EINSTEIN‟S EXPLANATION - PHOTOECLECTRIC EFFECTThe minimum amount photon of photon energy E = hf required to remove the least bound electron is called the E=0 work function (W) energy of electrons and has the units in an atom W hf o (work function) joules but eV are -ve more commonly - used. - -
  21. 21. EINSTEIN‟S EXPLANATION -PHOTOECLECTRIC EFFECTThe “least bound electron” is the electron in the outermost electron shell of the atom. This will be the easiest electron to pull away from the atom.It is also called the “most energetic electron” because all of the remaining energy given to it will be in the form of kinetic energy which will give it the highest speed of any of the released electrons.
  22. 22. EINSTEIN‟S EXPLANATION -PHOTOECLECTRIC EFFECTThe work function is given by: W = hfo fo = threshold frequencyThe threshold frequency is the minimum frequency required to free the “least bound” electron.
  23. 23. EINSTEIN‟S EXPLANATION - PHOTOECLECTRIC EFFECTWhen the photon falls on E = hf an electron with more energy than is needed to remove the bound (1 2 electron, the difference 2 m v ) m ax in energy is transformed into kinetic E = 0 energy of the electron.The least bound electron W hfo is also known as the -ve most energetic - electron. - -
  24. 24. EINSTEIN‟S EXPLANATION - PHOTOECLECTRIC EFFECT E = hfEincoming photon = K + Energy required for electron to escape (1 2For the least bound electron, 2 m v ) m axEincoming photon = K(max)+ W E = 0As E = hf W hfoEKmax = hf – hf0 -ve - - -
  25. 25. EXAMPLE 1A certain metal has a work function of 2.0eV. Will light of wavelength 4.0 x 10-7 m cause the ejection of photoelectrons and if so what will be their maximum velocity of ejection?
  27. 27. QUANTUM (MODERN) PHYSICSThe photon concept was used by Einstein to explain the experimental observations of the photoelectric effect.We will now go back to the observations made in the photoelectric effect experiment and look at Einstein’s explanation using quantum physics.
  28. 28. QUANTUM (MODERN) PHYSICSACTUAL QUANTUM OBSERVATION: EXPLANATION:The (kinetic) energy of A greater intensity means ejected that more photons will photoelectrons is fall on the surface. independent of the This will simply eject more intensity of radiation. electrons but NOT at a faster speed.
  29. 29. QUANTUM (MODERN) PHYSICSACTUAL QUANTUM OBSERVATION: EXPLANATION:Photoelectrons are The energy of a photon depends more likely to be on the frequency of radiation (E = hf). ejected by high A high-frequency photon has frequency than low more energy and so gives frequency radiation. more energy to the photoelectron. A high frequency photon is more likely to have greater energy than the work function.
  30. 30. QUANTUM (MODERN) PHYSICSACTUAL QUANTUM EXPLANATION: OBSERVATION: All of the energy of thePhotoelectrons are photon is given up to the ejected instantly. electron instantly. Experimental results show that the maximum time delay for the photoelectric effect is about 10-8s.
  31. 31. QUANTUM (MODERN) PHYSICSACTUAL QUANTUM OBSERVATION: EXPLANATION:A range of electron Once the work function is velocities of ejection subtracted, the remaining energy exists as kinetic are possible. energy. Depending on which electron absorbs the photon, varying amounts of kinetic energy may be left over.
  32. 32. QUANTUM (MODERN) PHYSICSSUMMARY – the photoelectric effect can be best explained using the Quantum Theory (light behaving as a particle) as opposed to the Classical Theory (light behaving as a wave).You would use the photoelectric effect in any question that asks you to prove that light behaves as particles (photons).
  33. 33. Experiment to test Einstein‟s modelThis is a diagram of an apparatus used to investigate the characteristics of photoelectric emission.It is used to try to determine Plank’s Constant (h) from E =hfMilikan used it to verify Einstein’s model in 1916
  34. 34. Experiment to test Einstein‟s modelThe cathode (negative) and anode (positive) are sealed in an evacuated glass tube to reduce the impedance (number of collisions) of the photoelectrons reaching the anode.When the light strikes the cathode it causes photoelectrons to be emitted.
  35. 35. Experiment to test Einstein‟s modelIf they cross the gap then they will create a current that will be read by a microammeter.The anode is made progressively more positive attracting more photoelectrons until the saturation current is reached.
  36. 36. Experiment to test Einstein‟s modelThis means that there cannot be more electrons given out from the cathode.It is attracting all of the electrons being given off at the cathode.
  37. 37. Experiment to test Einstein‟s modelNote that we DO NOT vary the frequency or the intensity during the time that we are making the anode more positive.During this time the current will get stronger, proof that the electrons are being emitted with different kinetic energies.
  38. 38. Experiment to test Einstein‟s modelOnly when you make the anode very positive do you finally attract the electrons that have very little kinetic energy (they are drifting around) due to the fact that they required a large amount of energy just to free them (their Work Function).
  39. 39. Experiment to test Einstein‟s modelIf the anode is made negative, electrons are repelled until there is no anode current. When the current is zero, the voltage applied is called the stopping voltage (Vs).
  40. 40. Experiment to test Einstein‟s modelAt this point even the most energetic electron (with the smallest work function and hence the most kinetic energy) will not be able to make it to the anode (due to repulsion).
  41. 41. Experiment to test Einstein‟s modelThis graph shows what happens as we change the frequency (colour) of the light and the voltage required to stop the most energetic electron for that particular frequency. Vs e = hf – hf0
  42. 42. The Photoelectric Effect1.The more intense the incident light, the greater the energy of ejection of the electron. The voltage required to repel the photoelectron from a current measuring device,should also be greater.
  43. 43. The Photoelectric Effect Intensity increased the number of ejected photoelectrons but, the energy is independent of the intensity of the radiation. The number of electrons emitted per second is,proportional to the intensity of the emitted light.
  44. 44. The Photoelectric Effect2.More photoelectrons should be ejected by low frequency radiation (i.e. red),than by high frequency radiation. Classical theory considers e-m waves to be,oscillating electric and magnetic fields.
  45. 45. The Photoelectric Effect Low frequency waves allow more time for the electron to move in one direction,before the field reverses and, the electron moves in the opposite direction.
  46. 46. The Photoelectric Effect High frequency waves would move so fast the electron would hardly begin to move in one direction, before it was forced to reverse direction, not ideal for ejection. Experiments showed that high frequency (UV) radiation ejected photoelectrons more readily, than low frequency.
  47. 47. The Photoelectric Effect There was a minimum frequency below which,no photoelectrons were ejected. This was called the threshold frequency, different for different materials.
  48. 48. The Photoelectric Effect 3.There should be a time delay between when a radiation is incident on a surface and,when the photoelectrons are ejected. Photoelectrons were ejected instantaneously.
  49. 49. The Photoelectric Effect4.The radiation‟s wavefront falls over the whole surface,billions of photoelectrons should be simultaneously ejected. By limiting the amount of light on a surface, a single electron could be ejected.
  50. 50. The Photoelectric Effect5.One velocity of ejection should be possible for radiation of one frequency. The electric current in the detector should immediately drop to zero once a critical voltage for repulsion of the single energy photoelectron is reached. Emitted photoelectrons have a range of ejection velocities and energies.
  51. 51. The Photoelectric Effect Complete the homework assignment On the wiki
  52. 52. WAVE NATURE OF MATTER It has been shown that in some circumstances, light exhibits certain behaviours characteristic of waves. In other circumstances, light behaves as particles. Could the reverse be true, namely that particles can behave as waves? This topic investigates this question.
  53. 53. WAVE NATURE OF MATTERDE BROGLIE’S HYPOTHESISCount Louis de Broglie (1892 - 1970) believed in the symmetry of nature. In 1923 he reasoned that if a photon could behave like a particle, then a particle could behave as a wave.
  54. 54. WAVE NATURE OF MATTER Special Relativity has given us the relationship E = mc2But, we also know E = hfThis gives hf = hc/λ = mc2 mc = h/ λ p = h/ λ
  55. 55. WAVE NATURE OF MATTERHe turned Compton‟s relationship to make wavelength the subject of the equation. Compton- “a photon has momentum” h p De Broglie- “An electron has a wavelength” h p
  56. 56. WAVE NATURE OF MATTERThis is called the de Broglie wavelength of a particle.All particles (electrons, protons, bullets, even humans) have a wavelength.They must be moving.They are called “matter waves”.
  57. 57. WAVE NATURE OF MATTERWe cannot see light. We can only make inferences about the nature of light by looking at its properties.Its properties indicate that it is both wave like and particle like in nature.
  58. 58. WAVE NATURE OF MATTERWe also cannot see atoms. We often think of them as exhibiting the properties of particles.But, because we have never seen them, could they be waves pretending to be particles?De Broglie suggested that particles, in some instances could be wave like.
  59. 59. EXAMPLE 1Calculate the de Broglie wavelength associated with a 1.0 kg mass fired through the air at 100 km/hr.
  61. 61. EXAMPLE 1 SOLUTIONNote the wavelength is so small that it cannot be detected and measured.We cannot create slits capable of diffracting such small wavelengths.Can a microscopic object give a more realistic wavelength?
  62. 62. EXAMPLE 2Calculate the de Broglie wavelength that would be associated with an electron accelerated from rest by a P.D. of 9.0V
  65. 65. EXAMPLE 3Calculate the de Broglie of a H atom moving at 158 m s-1 (interstellar space)
  66. 66. EXAMPLE 3 SOLUTION h mv -3 4 6 .6 2 6 x 1 0 -2 7 1 .6 7 2 x 1 0 x 158 = 2.50 x 10-9 mThese are X Rays which do not penetrate the atmosphere
  67. 67. An EXPERIMENT to verify de BroglieC.J. Davisson and L.H. Germer performed an experiment in 1927 to verify de Broglie‟s hypothesis.
  68. 68. DAVISSON-GERMEREXPERIMENTElectrons were allowed to strike a nickel crystal. The intensity of the scattered electrons is measured for various angles for a range of accelerating voltages.
  70. 70. DAVIDSON-GERMEREXPERIMENTIt was found that a strong „reflection‟ was found at θ = 50 when V = 54V.This appeared to be a place of constructive interference, suggesting that the “matter waves” from the electrons were striking the crystal lattice and diffracting into an interference pattern.
  71. 71. DAVIDSON-GERMEREXPERIMENTThe interatomic spacing of Nickel is close to the „wavelength‟ of an electron. Therefore it would seem possible that electron matter waves could be diffracted.Davidson and Germer set out to verify that the electrons were behaving like a wave using the following calculations.
  72. 72. DAVIDSON-GERMEREXPERIMENTTheoretical Result (according to de Broglie’s calculation)The kinetic energy of the electrons is 1/2 mv2 = Ve
  73. 73. DAVIDSON-GERMEREXPERIMENTThe de Broglie wavelength is given by: h h mv 2VemFor this experiment: -34 6 . 625 x 10 -19 - 31 2 x 54 x (1.6 x 10 ) x (9.11 x 10 ) -10 1 . 67 x 10 m
  74. 74. DAVIDSON-GERMEREXPERIMENTExperimental Result (according to Davidson- Germer)X-ray diffraction had already shown the interatomic distance was 0.215 nm for nickel.Since θ = 50 , the angle of incidence to the reflecting crystal planes in the nickel crystal is 25 as shown below:
  76. 76. DAVIDSON-GERMEREXPERIMENT dsin θ = mλFor the first order reinforcement… λ = dsinθ = (.215 x 10-9)(sin50 ) = 1.65 x 10-10 m
  77. 77. DAVIDSON-GERMEREXPERIMENTThe close correspondence between the theoretical prediction for the wavelength by de Broglie (1.67 x 10-10 m) and the experimental results of Davidson- Germer (1.65 x 10-10 m) provided a strong argument for the de Broglie hypothesis.
  78. 78. SPECTRASpectra provide a “picture or model” of what the atom looks like.About midway through the 19th century, it was discovered that if a gas could be stimulated to produce light, the colour could indicate the nature of the gas.The light was passed through a prism or diffraction grating to produce a spectrum.
  79. 79. SPECTRAThere are 3 types of spectrum:CONTINUOUS SPECTRUMHeating an object such as a filament in an incandescent globe produces this. When viewed through a spectroscope, all spectral colours are seen.An object‟s spectrum can extend beyond the visible part of the spectrum.
  80. 80. CONTINUOUS SPECTRAA hotter object will produce more energy at all wavelengths than does a cooler object.The hotter the object, the more energy is emitted at shorter wavelengths. The distribution of wavelengths depends on the temperature.This is why the colour of objects change as the object is heated.
  82. 82. LINE EMISSION SPECTRAProduced from light emitted by excited, or energised gases.When gaseous atoms become excited, they emit characteristic colours (sodium - yellow and mercury - mauve).
  83. 83. LINE EMISSION SPECTRALooking through a spectrometer, the light is made up of specific frequencies and is referred to as spectral lines and the full spectrum seen is known as a line emission spectrum.Unlike solids and dense liquids, gases do not produce a continuous spectrum.
  84. 84. LINE EMISSION SPECTRAThey are not vibrating back and forth at a variety of frequencies because they are not bonded together.Instead, they emit SPECIFIC frequencies of light as seen in the spectroscope.
  85. 85. LINE EMISSION SPECTRAThese specific frequencies are produced by electrons falling from higher energy levels to lower energy levels.A photon of a specific frequency is emitted in the process.
  87. 87. ABSORPTION SPECTRAThis is a continuous spectrum with dark lines. This means colours are removed due to their absorption by matter.In 1814, a German scientist, Fraunhofer, focused the spectrum of light from the sun and found over 700 dark lines or gaps in the spectrum.
  89. 89. ABSORPTION SPECTRAIf we can match the absorption spectrum of the sun with the absorption spectra of an element like calcium, then we can say that calcium is in the sun‟s outer atmosphere.
  90. 90. ABSORPTION SPECTRAThe absorbing material used in a lab is generally a gas or liquid but any state could be used. The dark lines depend on the nature of the absorbing material.The absorption spectrum is as much a fingerprint of the element giving off the E-M energy as is the emission spectrum.
  91. 91. ABSORPTION SPECTRAIn all cases the absorption and the emission spectra will match perfectly.
  93. 93. THE BOHR MODEL OF THEATOMIn 1911 Bohr ignored all the previous descriptions of the electronic structure as they were based on classical physics.This allowed the electron to have any amount of energy.Planck and Einstein used the idea of quanta for the energy carried by light.
  94. 94. THE BOHR MODEL OF THEATOMBohr assumed that the energy carried by an electron was also quantized.From this assumption, he formed three postulates (good intelligent guesses) from which he developed a mathematical description.
  95. 95. THE BOHR MODEL OF THEATOM Bohr atom + free e - 0 - } bound e - energy levels
  96. 96. THE BOHR MODEL OF THEATOMIn summary, if the atom had electrons that varied in their energy levels, you would expect to get random frequencies emitted.This is not the case.Electrons give off photons of SPECIFIC frequencies.More evidence for the Quantum Theory!
  97. 97. THE BOHR MODEL OF THEATOMAn electron can be moved to a higher energy level by… 1. INCOMING PHOTON- Must be of exactly the same energy as E2 – E1 2. INCOMING ELECTRON- remaining energy stays with the incoming electron. 3. HEAT- gives the electron vibrational energy.
  98. 98. THE BOHR MODEL OF THEATOM IONIZATION- energy required to remove an electron from the atom. Example: the ionization energy required to remove an electron from its ground state (K=1) for Hydrogen is 13.6 eV.
  99. 99. THE BOHR MODEL OF THE ATOMLet’s prove this using the 2 ionized atom (electron unbound and free to takeany energy) n= n=5 following formula… -0.85 0 n = 4 (N shell) -1.51 n =3 (M shell) LYMAN SERIES -2 Paschen series -3.4 n = 2 (L shell)En-Em = hf -4 Balmer series 1st excited state energy (eV)(13.6 –0.85)(1.6X10-19)=hc/λ -6λ=9.7X10-8 m (98nm) -8This is UV Light -10 -12 -13.6 n = 1 (K shell) -14 Lyman series ground state
  100. 100. THE BOHR MODEL OF THE ATOM BALMER SERIES 2En-Em = hf ionized atom (electron unbound and free to takeany energy) n= n=5 0(3.4 –0.85)(1.6X10-19)=hc/λ -0.85 -1.51 -2 Paschen series n = 4 (N shell) n =3 (M shell) λ=4.87X10-7 m -3.4 -4 n=2 Balmer series 1st excited state (L shell) energy (eV) = 487 nm -6 -8This is Blue Light. -10 -12 -13.6 n=1 (K shell) -14 Lyman series ground state
  101. 101. EXAMPLE 1Use the first three energy levels for the electron in hydrogen to determine the energy and hence wavelength of the lines in its line emission spectrum. -1.51 eV n=3 -3.4 eV n=2 -13.6 eV n=1
  102. 102. EXAMPLE 1 SOLUTIONFrom the diagram, the atom can be excited to the first (n = 2) and second (n = 3) excited states. From these, it will return to the ground state emitting a photon. The electron can make the following transitions:
  103. 103. EXAMPLE 1 SOLUTIONn = 3 n = 1, Ephoton = E3 - El = -1.51 - (-13.6) = 12.09eVn = 2 n = l, Ephoton = E2 - E1 = -3.4 - (-13.6) = 10.2eV
  104. 104. EXAMPLE 1 SOLUTIONn = 3 n = 2, Ephoton = E3 - E2 = -1.51 - (-3.4) = 1.89 eV n=3 n=2 n=1
  105. 105. EXAMPLE 1 SOLUTIONTo find the wavelengths of the three photons, use hc hc E = hf ie. = ENote: convert eV to J
  106. 106. EXAMPLE 1 SOLUTION -34 -8 n=3 n=l, 6.6 x 10 x 3 x 10 -19 12.09 x 1.6 x 10 = 1.024 x 10 -7 m = 102 nm (ultraviolet)
  107. 107. EXAMPLE 1 SOLUTIONn=2 n=l, λ= 6 . 6 x 10 34 x 3 x 10 8 10.2 x 1.6 x 10 - 19 = 1.21 x 10-7 m = 121nm (ultraviolet)n=3 n=2, λ= 6 . 6 x 10 34 x 3 x 10 8 1.89 x 1.6 x 10 - 19 = 6.55 X 10-7m = 655 nm (visible-red)
  108. 108. EXAMPLE 1 SOLUTIONNote that we have two n=3 Lyman series lines (those ending in the ground state) and one Balmer line n=2 (ending in the first excited state). n=1
  109. 109. EXAMPLE 1 SOLUTIONThose from the Lyman series produce lines n = 3 in the ultra-violet part of the spectrum while the line in the Balmer n = 2 series produces a line in the visible part of the spectrum. n=1
  110. 110. EXAMPLE 2The valence (outer) electron of sodium is involved in the emission of two photons that have almost the same energy. The wavelengths are 589.0 nm and 589.6 nm. It is known that the valence electron falling to the ground state from one or two higher levels will produce these photons. Determine the excitation energies of these two energy levels, and find the difference in their energy. (Work to 4 sig. Figs.)
  111. 111. EXAMPLE 2 SOLUTIONThe diagram represents the two energy levels whose energies are El and E2 above the ground state. E1 E2 ground state
  112. 112. EXAMPLE 2 SOLUTION -34 8 hc 6.625 x 10 x 3.00 x 10E1 hf 1 -7 1 5.896 x 10  = 3.371 X 10-19 J hc 6.625 x 10 -34 x 3.00 x 10 8 E2 -7 2 -19 5.890 x 10  = 3.374 X 10 J
  113. 113. EXAMPLE 2 SOLUTIONNow it is more convenient to work in electron volts. E1 = 2.107eV, and E2 = 2.109eV, so the energy difference is 0.002 eV (3.2 x 10-22 J). The energy level diagram can now be drawn. E1 = 2.109 eV E 2 = 2.107 eV ground 589.0 nm 589.6nm state
  114. 114. Electron in a box modelThe electron is bound to the nucleus by the Coulombic attractionConsider, the electron to be confined in a one dimensional box whose edges are defined by 1/rClassical wave theory states that such a confined wave would be a standing wave and hence λn =2 L /n …where n = 1,2,3… But pn = h / λn = n h / 2 LAlso EK = p2/2m … En = n2h2/ 8mL2
  115. 115. Schrödinger ModelThe Bohr Model was a landmark in the history of Physics Limitations show that a new model was neededLess than 2 years after de Broglie gave us the matter wave Erwin Schrödinger (1887-1961) an Austrian Physicist developed a new comprehensive theory
  116. 116. Schrödinger Model
  117. 117. Schrödinger ModelDe Broglie determined the wavelength and momentum of a matter waveWhat about amplitude?The amplitude of a matter wave is given the symbol Greek letter psi
  118. 118. Schrödinger ModelSchrödinger developed an equation to determine the wavefunction, . The wavefunction represents the amplitude of a matter wave as a function of time and position It is a differential equation – the solution of which exactly predicts the line spectra of a hydrogen atom may vary in magnitude from point to point in space and time
  119. 119. Schrödinger ModelConsider Young‟s Double Slit Experiment If slits are in the order of the wavelength of light An interference pattern would be seen Reduce the flow of photons (or electrons) to one at a time
  120. 120. Schrödinger ModelInitially electrons appears to be random
  121. 121. Schrödinger ModelIf time is allowed to elapse, the pattern starts to become visible
  122. 122. Schrödinger ModelEventually the pattern follows that expected by wave theory
  123. 123. Schrödinger ModelWe actually see:
  124. 124. Schrödinger ModelWhere 2 is zero;  A minimum in the pattern would be seenWhere 2 is a maximum  A maximum in the pattern would be seen
  125. 125. Schrödinger ModelTo get an interference pattern;  Electrons must pass through both slits at the same timeThis is possible as an electron is as much a wave as a particleWhat would happen if we covered up one slit?
  126. 126. Schrödinger ModelElectron would pass through one slit only;  A diffraction pattern would be seenIf we then covered up the other slit?  A different diffraction pattern would be seenThere would be no interference pattern
  127. 127. Schrödinger ModelIf both slits were open  Electron passes through both slits  As if it were a wave  Forming an interference patternYet each electron would make a tiny spot on the screen as if it were a particle
  128. 128. Schrödinger ModelThe main point: Treat electrons as waves;  represents the wave amplitude Treat electrons as particles  Must treat them on probabilistic basis 2 gives the probability of finding a given electron at a pointWe cannot predict or follow the path of an electron precisely through space and time
  129. 129. Schrödinger ModelCan use Schrödinger‟s equation to determine the probability of finding an electron at any given place or time around a nucleus 2 tellsus an electron is more likely to be found close to the nucleus than far awayThis allowed the development of an electron cloud
  130. 130. Heisenberg UncertaintyPrincipleIn 1927 Werner Heisenberg proposed a principle that helped understand the interpretation of the wavefunction. Let the uncertainty in position be Δx Let the uncertainty in momentum be ΔpThe uncertainty principle states Δx Δp h (More accurately h/4 )
  131. 131.  If p is known precisely then = h/p is also known But a completely defined means that a wave must be infinite in space and time Therefore if p is known then we cant know the position and vice versa Uncertainty principle also applies to energy ΔE Δt hThis explains why spectral lines are of finite width