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The Wave – Particle Duality

The Wave – Particle Duality. OR. Light Waves. Until about 1900, the classical wave theory of light described most observed phenomenon. Light waves: Characterized by: Amplitude (A) Frequency ( n ) Wavelength ( l ) Energy a A 2. And then there was a problem….

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The Wave – Particle Duality

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  1. The Wave – Particle Duality OR

  2. Light Waves Until about 1900, the classical wave theory of light describedmost observed phenomenon. • Light waves:Characterized by: • Amplitude (A) • Frequency (n) • Wavelength (l) Energy a A2

  3. And then there was a problem… However, in the early 20th century, several effects were observed which could not be understood using the wave theory of light.Two of the more influential observations were:1) The Photo-Electric Effect 2) The Compton Effect I will describe each of these today…

  4. What if we try this ? Vary wavelength, fixed amplitude electrons emitted ? No No Yes, withlow KE No Yes, withhigh KE No No Photoelectric Effect (I) “Classical” Method Increase energy by increasing amplitude electrons emitted ? No electrons were emitteduntil the frequency of the light exceeded a critical frequency, at which point electrons were emitted from the surface! (Recall: small l large n)

  5. Photoelectric Effect (II) • Electrons are attracted to the (positively charged) nucleus by theelectrical force • In metals, the outermost electrons are not tightly bound, and canbe easily “liberated” from the shackles of its atom. • It just takes sufficient energy… Classically, we increase the energyof an EM wave by increasing theintensity (e.g. brightness) Energy a A2 But this doesn’t work ??

  6. PhotoElectric Effect (III) • An alternate view is that light is acting like a particle • The light particle must have sufficient energy to “free” theelectron from the atom. • Increasing the Amplitude is just simply increasing the numberof light particles, but its NOT increasing the energy of each one! Increasing the Amplitude does diddly-squat! • However, if the energy of these “light particle” is related to their frequency, this would explain why higher frequency light canknock the electrons out of their atoms, but low frequency light cannot…

  7. Photo-Electric Effect (IV) • In this “quantum-mechanical” picture, the energy of thelight particle (photon) must overcome the binding energy of the electron to the nucleus. • If the energy of the photon does exceed the binding energy, theelectron is emitted with a KE = Ephoton – Ebinding. • The energy of the photon is given by E=hn, where theconstant h = 6.6x10-34 [J s] is Planck’s constant. “Light particle” Before Collision After Collision

  8. Photons • Quantum theory describes light as a particle called a photon • According to quantum theory, a photon has an energy given byE = hn = hc/lh = 6.6x10-34 [J*sec]Planck’sconstant, after the scientist Max Planck. • The energy of the light is proportional to the frequency, and inversely proportional to the wavelength! The higher the frequency (lower wavelength) the higher the energy of the photon! • 10 photons have an energy equal to ten times a single photon. • The quantum theory describes experiments to astonishing precision, whereas the classical wave description cannot.

  9. The Electromagnetic Spectrum Shortest wavelengths (Most energetic photons) E = hn = hc/l h = 6.6x10-34 [J*sec](Planck’s constant) Longest wavelengths (Least energetic photons)

  10. Vary wavelength, fixed amplitude electrons emitted ? No E1 = hn1 IncreaseEnergy Yes, withlow KE E2 = hn2 IncreaseEnergy Yes, withhigh KE E3 = hn3 Interpretation of Photoelectric Effect E3 > E2 > E1 Photoelectric Effect Applet

  11. Incident X-raywavelength l1 M A T T E R Scattered X-raywavelength l2 l2 >l1 e Electron comes flying out The Compton Effect In 1924, A. H. Compton performed an experiment where X-rays impinged on matter, and he measured the scattered radiation. Louis de Broglie Problem: According to the wave picture of light, the incident X-ray gives up energy to the electron, and emerges with a lower energy (ie., the amplitude is lower), but must have l2=l1.

  12. Electroninitially atrest Scattered X-ray p2 = h / l2 Incident X-rayp1 = h / l1 l2 >l1 e e pe e Quantum Picture to the Rescue • If we treat the X-ray as a particle with zero mass, and momentum p = E / c, everything works ! Compton found that if the photon was treated like a particlewith mometum p=E/c, he could fully account for the energy & momentum (direction also) of the scattered electron and photon! Just as if 2 billiard balls colliding!

  13. DeBroglie’s Relation p = h / l • The smaller the wavelength the larger the photon’s momentum! • The energy of a photon is simply related to the momentum by: E = pc (or, p = E / c ) • The wavelength is related to the momentum by: l = h/p • The photon has momentum, and its momentum is givenby simply p =h / l .

  14. Momentum of Photons If I have a photon with energy E=1 [GeV], what is its momentum? p = E / c = (1 [GeV])/c = 1 [GeV/c] … That’s it! If I have a photon with momentum 5 GeV/c, what is its energy? E = pc = (5 GeV/c) * c = 5 [GeV] … whallah ! So, the only difference between a photons’ energy and momentum is: Energy [GeV] momentum [GeV/c]Don’t forget though that the “c” in [GeV/c] really means 3x108 [m/s].

  15. Scattering Problem Electroninitially atrest Incident X-raywavelength li=1.5 [nm] lf e e KE=0.2 [keV] Before After • Compute the energy of the 1.5 [nm] X-ray photon. E = hc/l = (6.6x10-34 [J s])(3x108 [m/s]) / (1.5x10-9 [m]) = 1.3x10-16 [J]

  16. Scattering Example (cont) • Express this energy in [keV]. 1.3x10-16 [J] * (1 [eV] / 1.6 x10-19 [J]) = 825 [eV] = 0.825 [keV] • What is the magnitude of the momentum of this photon? p = E / c = 0.825 [keV]/ c = 0.825 [keV/c] • After the collision the electron’s energy was found to be 0.2 [keV]. What is the energy of the scattered photon? A) 0.2 [keV] B) 0.625 [keV] C) 1.025 [keV] D) 0.825 [keV] Since energy must be conserved, the photon must have E=0.825-0.2 = 0.625 [keV] • What would be the wavelength of the scattered photon? HW exercise !

  17. Summary of Photons • Photons can be thought of as “packets of light” which behave as a particle. • To describe interactions of light with matter, one generally has to appeal to the particle (quantum) description of light. • A single photon has an energy given by E = hc/l, where h = Planck’s constant = 6.6x10-34 [J s] and, c = speed of light = 3x108 [m/s]l = wavelength of the light (in [m]) • Photons also carry momentum. The momentum is related to the energy by: p = E / c = h/l

  18. Matter Waves ? One might ask: “If light waves can behave like a particle, might particles act like waves”? The short answer is YES. The explanation lies in the realm of quantum mechanics, and is beyond the scope of this course. However, you already have been introduced to the answer.Particles also have a wavelength given by: l = h/p = h / mv • That is, the wavelength of a particle depends on its momentum, just like a photon! • The main difference is that matter particles have mass, and photons don’t !

  19. Matter Waves (cont) Compute the wavelength of a 1 [kg] block moving at 1000 [m/s]. l = h/mv = 6.6x10-34 [J s]/(1 [kg])(1000 [m/s]) = 6.6x10-37 [m].  This is immeasureably small. So, on a large scale, we cannot observe the wave behavior of matter Compute the wavelength of an electron (m=9.1x10-31 [kg]) moving at 1x107 [m/s]. l = h/mv = 6.6x10-34 [J s]/(9.1x10-31 [kg])(1x107 [m/s]) = 7.3x10-11 [m]. This is near the wavelength of X-rays

  20. Electron Microscope • The electron microscope is a device which uses the wave behavior of electrons to make images which are otherwise too small using visible light! This image was taken with a Scanning Electron Microscope (SEM). These devices can resolve features downto about 1 [nm]. This is about 100 times better than can be done with visible light microscopes! IMPORTANT POINT HERE: High energy particles can be used to reveal the structure of matter !

  21. Remarks on Particle Probes • We have now asserted that high energy particles (electrons in thecase of a SEM) can provide a way to reveal the structure of matterbeyond what can be seen using an optical microscope. • The higher the momentum of the particle, the smaller thedeBroglie wavelength. • As the wavelength decreases, finer and finer details about thestructure of matter are revealed ! • We will return to this very important point.  To explore matter at its smallest size, we need very high momentum particles!  Today, this is accomplished at facilities often referred to as “atom-smashers”. We prefer to call them “accelerators”  More on this later !

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