as physics

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ghoran
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245463
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as physics
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2013-12-24 09:29:22
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unit one section two electromagnetic radiation quantum phenomena
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  1. most of the time we think of light as a wave , but there are some situations where it acts as a particle too . the most famous of these cases is the
    photoelectric effect
  2. what is the photoelectric effect
    • if you shine radiation  of a high enough frequency onto the surface of a metal , it will instantly emit electrons . for most metals , this frequency falls in the u.v. range . 
    • Because of the way atoms are bonded together in metals , metals contain a sea of free de-localised electrons that are able to move about the metal . The free electrons on or near the surface of the metal absorb energy from the radiation , making them vibrate .
    • if an electron absorbs enough energy , the bonds holding it to the metal break and the electron is released . This is called the photoelectric effect and the electrons emitted are called photoelectrons
  3. draw a diagram to show the photoelectric effect
  4. what are the main conclusion from the photoelectric effect
    • 1) for a given metal no photoelectrons are emitted if the radiation has a frequency below a certain value called threshold frequency 
    • 2) the photoelectrons are emitted with a variety of kinetic energies ranging from zero to some maximum value . this maximum kinetic energy increases with the frequency of the radiation 
    • 3) the intensity of radiation is the amount of energy per second hitting an area of the metal 
    • 4) the number of photoelectrons emitted per second is proportional to the intensity of the radiation
  5. the key thing about the photoelectric effect is that it shows that
    light can't just act as a wave . Certain observations of the photoelectric effect can't be explained by classic wave theory
  6. wave theory says that for a particular frequency of EM wave the energy carried should be proportional to
    the intensity of the beam . The energy carried by the Em wave would also be spread evenly over the wave front
  7. if we assumed that light was only a wave and used wave theory , if we shone an EM wave on a metal
    each free electron on the surface of the metal would gain a bit of energy from each incoming wave . Gradually each electron would gain enough energy to leave the metal . If the EM wave had a lower frequency i.e. (was carrying less energy) it would take longer for the electrons to gain enough energy but it would happen eventually . However electrons are never emitted unless the wave is above a threshold frequency - so wave theory can't explain the threshold frequency
  8. kinetic energy of photoelectrons
    • the higher the intensity of the wave , the more energy it should transfer to each electron - the kinetic energy should increase with intensity
    • wave theory can't explain the fact that kinetic energy depends only on the frequency in the photoelectric effect  
  9. what did Max Planck suggest
    he was the first to suggest that EM waves can only be released in discrete packets or quanta
  10. the energy , E , carried by one of these wave-packets is
    • E = energy of one packet in J 
    • h = Planck constant = 6.63 x 10^-34 J
    • f = frequency of light in Hz
    • c = speed of light in a vacuum = 3.00 x 10^8 m/s 
    • other symbol = wavelength in m
  11. Einstein's photons
    • Einstein went further by suggesting that EM waves (and the energy they carry) can only exist in discrete packets . He called these wave packets photons . 
    • he saw these photons of light as having a one-on-one particle like interaction with an electron in a metal surface . Each photon would transfer all its energy to one specific electron . the photon model could be used to explain the photoelectric effect
  12. the photon model of light can explain the observations and conclusions for the
    photoelectric effect that the wave model of light can't
  13. when EM radiation hits a metal , the metal is bombarded by
    photons
  14. if one of these photons collides with a free electron , the electron will gain energy equal to
    hf (as E = hf)
  15. before an electron can leave the surface of the metal , it needs enough energy to break the bonds holding it there . this energy is called the
    work function energy (symbol ø) and its value depends on the metal .
  16. if the energy gained from a photon is greater than the work function energy
    the electron is emitted . If it isn't , the electron will just shake about a bit , the release the energy as another photon . The metal will heat up but no electrons will be emitted . Since , for electrons to be emitted hf must be equal or greater than or equal to the work function energy . the threshold frequency must be equal to the work functyion energy/h 
  17. the energy transferred from EM radiation to an electron is
    the energy it absorbs from one photon , hf .the kinetic energy is the EM radiation will be carrying when it leaves the metal is hf - any other energy losses . these energy looses are the reason the electrons emitted from a metal have a range of kinetic energies 
  18. the minimum amount of energy an electron can loose is the 
    work function energy , so the maximum kinetic energy , Ek , is given by the equation Ek = hf - the work function energy . rearranging this equation gives you the photoelectric effect equation 
  19. kinetic energy = 
    1/2 mv2
  20. this means the maximum kinetic energy a photoelctron can have is
    E= 1/2mv2

    • m = mass of an electron = 9.11 x 10-31 
    • v = maximum velocity of an electron emitted
  21. you can use this to write out the photoelectric equation as
    hf = work function energy + 1/2 mv2
  22. the kinetic energy of the electrons is
    independent of the intensity because they can only absorb one photon at a time
  23. the energies of electrons in an atom are usually so tiny that it makes sense to use a more appropriate unit than the joule . The electron volt is defined as
    the kinetic energy carried by an electron after it has been accelerated through a potential difference of 1 volt
  24. the energy gained by an electron is equal to
    to the accelerating voltage
  25. you can convert between eV and J with this formula
    1eV = 1.6 x 10-19 J
  26. electrons in an atom can only exist
    in certain well-defined energy levels .
  27. each energy level is given a number , with n=1 representing the
    lowest energy level an electron can be in - the ground state
  28. we say an electron is excited when
    one or more of its electrons is in an energy level higher than the ground state
  29. electrons can move down an energy level by
    emitting a photon . Since these transitions are between a definite energy levels , the energy of each photon emitted can only take a certain allowed value .
  30. :)
  31. the energy carried by a photon emitted after a transition is equal to
    the difference in energies between the two levels of the transition
  32. electrons can also move up energy levels if
    • they absorb a photon with the exact energy difference between the two levels
    • ΔE must equal hf
  33. the movement of an electron to a higher energy level is called
    excitation
  34. ΔE =
    Ex - E
  35. ionization
    when an electron has been removed from an atom the atom has been ionized .
  36. The energy of each energy level within an atom shows the
    amount of energy needed to remove an electron from that level
  37. the ionization energy of an atom is
    the amount of energy needed to remove an electron from the ground state atom
  38. what do Fluorescent tubes use to emit light
    the excitation of electrons and photon emission . They contain mercury vapor , across which a high voltage is applied . When fast moving electrons (emitted by electrodes in the tube and accelerated by high voltage) collide with the electrons in the mercury atoms . the atomic mercury electrons are excited to a higher energy level . when these excited electrons return to their ground states , they lose energy by emitting high energy photons in the UV range . the photons emitted have a range of energies and wavelengths that correspond to the different transitions of the electrons . a phosphorous coating on the inside of the tube absorbs these photons exciting its electrons to much higher energy levels . These electrons then cascade down the energy levels and lose energy by emitting many lower energy photons of visible light
  39. if you split the light from a fluorescent tube with a prism or a diffraction grating you get a
    • line spectrum . a line emission spectrum is seen as a series of bright lines against a black background . each line corresponds to a particular wavelength of light emitted by the source 
  40. line spectra provide evidence that 
    the electrons in atoms exist in discrete energy levels . atoms can only emit photons with energy energies equal to the difference between the two energy levels . since only certain photon energies are allowed , you only see the corresponding wavelengths in the line spectrum  
  41. the spectrum of white light is 
    • continuous . if you split the light up with a prism , the colours all merge into each other - there aren't any gaps in the spectrum . Hot things emit a continuous spectrum in the visible and infra-red
    •  
  42. draw a diagram to show the continuous spectrum of white light , the emission lines of white light and the absorption lines of white light
    • <--------------------- decreasing wavelength
  43. what are line absorption spectra
    you get a line absorption spectrum when a continuous spectrum of energy (white light) passes through  a cool gas . at low temperatures , most of the electrons in the gas atoms will be in their ground state . Photons of the correct wavelength are absorbed by electrons to excite them to higher energy levels . these wavelengths are then missing from the continuous spectrum when it comes out of the other side of the gas . you see a continuous spectrum with black lines in it corresponding to the absorbed lines 
  44. if you compare the absorption and emission spectra of a particular gas ,
    the black lines in the absorption spectrum match up to the bright lines in the emission spectrum 
  45. the gas in the absorption spectra is ....
    the gas in the emission spectra is .....
    • cool
    • excited 
  46. when a beam of light passes through a narrow gap , it spreads out . this is called 
    diffraction
  47. diffraction can only be explained using 
    waves . If light was acting as a particle , the light particles in the beam would either not get through the gap (if they were too big) or just pass straight through and the beam would be unchanged 
  48. the results of photoelectric effect experiments can only be explained by
    thinking of light as a series of particle like photons . If a photon of light is a discrete bundle of energy , then it can interact with an electron in a one to one way . all the energy in the photon is given to one electron 
  49. the photoelectric effect and diffraction show that light behaves as both a particle and a wave this is known as 
    wave particle duality 
  50. Louis De Broglie made a bold suggestion in his PhD thesis . He said that if wave like light showed particle properties , 
    particles like electrons should be expected to show wave like properties
  51. the De Broglie equation relates
    a wave property (wavelength) to a moving particle property (momentum)
  52. the De Broglie equation is 
  53. The De Broglie wave of a particle can be interpreted as 
    a probability wave . Many physicists at that time weren't very impressed his ideas were just speculation . But later experiments confirmed the wave nature of electrons and other particles 
  54. diffraction patterns are observed when
    accelerated electrons in a vacuum tube interact with the spaces in a graphite crystal 
  55. what proves electrons have wave like properties 
    diffraction patterns are observed when accelerated electrons in a vacuum tube interact with the spaces in a graphite crystal . As electrons pass through the spaces they diffract just like waves passing through a narrow slit and produce a pattern of rings
  56. according to the wave theory , the spread of lines in the diffraction pattern increases  
    if the wavelength of the waves is greater . 
  57. in electron diffraction experiments a smaller accelerating voltage , i.e. slower electrons gives 
    widely spaced rings .
  58. increase the electron speed and the diffraction pattern circles 
    squash together towards the middle . This fits in with the De Broglie equation  - if the velocity is greater the wavelength is shorter and the spread of the lines is smaller 
  59. in general wavelength for electrons accelerated in a vacuum tube is about the same size as 
    electromagnetic waves in the Xray part of the spectrum 
  60. you only get diffraction if a particle
    interacts with an object about the same size as its De Broglie wavelength . so you only get electrons acting as a wave if the electron interacts with an object the same size as the De Broglie wavelength of the electron
  61. a tennis ball , with a mass of 0.058kg and a speed of 100ms-1 has a De Broglie wavelength of 10-34m . that's 10-19 times smaller than the nucleus of an atom and there's
    nothing that small for it to interact with , and so it acts as a particle
  62. an electron of mass 9.11 x 10-31 kg is fired from an electron gun at 7 x 106 ms-1 . what size object will the electron need to interact with in order to diffract
    • an electron will diffract when the size of the object is roughly the same size as it's De Broglie wavelength , so we need to find the wavelength
    • momentum of electron = mv
    • 9.11x10-31 x 7x106 = 6.377 x 10-24 kgms-1 
    • substitute this into the De Broglie's equation 
    • wavelength = h/mv
    • 6.63x10-34/6.377x10-24 = 1x10-10 m (2sf) so only crystals with an atom layer spacing around this size are likely to cause the diffraction of this electron  
  63. electrons with a wavelength of 1.7x10-10m are diffracted as they pass between atoms in a crystal lattice . calculate the velocity of the electrons 
    • substitute wavelength = 1.7x10-10m and h=6.63x10-34 js and m=9.11x10-31 kg into the De Broglie equation 
    • wavlength = h/mv 
    • 1.7x10-10 = 6.63x10-34/9.11x10-31 x v 
    • rearrange and solve for v
    • v = 6.63x10-34/9.11x10-31 x 1.7x10-10 = 4,280,000 ms-1 (to 3sf)
  64. a shorter wavelength gives
    less diffraction effects 
  65. a shorter wavelength gives less diffraction effects . this fact is used in the electron
    microscope 
  66. diffraction effects ..... .... on an image 
    blur detail 
  67. if you want to resolve a tiny detail in an image , you need a 
    shorter wavelength
  68. Light blurs out detail more than 
    electron waves do , so an electron microscope can resolve finer detail than a light microscope . they can let you look at things as tiny as a single strand of DNA

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