Slide 1 / 83 Quantum Theory and Atomic Models
Slide 2 / 83 The Electron · Streams of negatively charged particles were found to emanate from cathode tubes. · J. J. Thompson is credited with their discovery (1897).
Slide 3 / 83 Discovery and Properties of the Electron It was found that these rays could be deflected by electric or magnetic fields. By adjusting those fields the charge to mass ratio of the unknown "ray" was found.
Slide 4 / 83 Discovery and Properties of the Electron First, they found the velocity of the particle by adjusting the magnetic field electric forces so that they cancelled out; the "ray" traveled in a straight line. Σ F= ma F B - F E = 0 F E v F B qvB = qE v = E/B Since they could measure E and B, they could calculate v.
Slide 5 / 83 Discovery and Properties of the Electron Then they turned off the electric field and the particle moved in a circular path. They measured the radius of the circle, by seeing where the particle struck the tube, and then determined the charge to mass ratio: q/m. Σ F= ma F B = ma v F B qvB = mv 2 /r qB = mv/r r q/m = v/Br q/m = 1.76 x 10 11 C/kg
Slide 6 / 83 1 Which one of the following is not true concerning cathode rays? A They originate from the negative electrode. B They travel in straight lines in the absence of electric or magnetic fields. C They impart a negative charge to metals exposed to them. D They are made up of electrons. E The characteristics of cathode rays depend on the material from which they are emitted.
Slide 7 / 83 Millikan Oil Drop Experiment Once the charge/mass ratio of the electron was known, determination of either the charge or the mass of an electron would yield the other.
Slide 8 / 83 Discovery and Properties of the Electron The mass of each droplet was estimated by its size. The electric field was adjusted so the drop fell with constant velocity. The data showed that the charge was always an integral multiple of a smallest charge, e. That must be the charge of one electron. Σ F= ma F E - mg = 0 qE = mg q = mg/E F E mg q was always an integer multiple of the same number, which was given the symbol "e"
Slide 9 / 83 2 Which of these could be the charge of an object? 0.80 x 10 -19 C A 2.0 x 10 -19 C B 3.2 x 10 -19 C C 4.0 x 10 -19 C D
Slide 10 / 83 3 The charge on an electron was determined in the __________. A cathode ray tube, by J. J. Thompson B Rutherford gold foil experiment C Millikan oil drop experiment D Dalton atomic theory E atomic theory of matter
Slide 11 / 83 Blackbody Radiation All objects emit electromagnetic radiation which depends on their temperature: thermal radiation. A black body absorbs all electromagnetic radiation (light) that falls on it. Because no light is reflected or transmitted, the object appears black when it is cold. However, black bodies emit a temperature-dependent spectrum termed blackbody radiation.
Slide 12 / 83 Blackbody Radiation At normal temperatures, we are not aware of this radiation. But as objects become hotter, we can feel the infrared radiation or heat. At even hotter temperatures, objects glow red and at still hotter temperatures, object can glow white hot such at the filament in a light bulb.
Slide 13 / 83 Blackbody Radiation This figure shows blackbody radiation curves for three different temperatures. The wavelength at the peak, # p , is related to the temperature by: # p T = 2.90 x 10 -3 m-K Classical physics couldn't explain the shape of these spectra.
Slide 14 / 83 4 Which of the following colors would indicate the hotest temperature? A Black B Red C Yellow D Blue
Slide 15 / 83 Planck’s Quantum Hypothesis · The wave nature of light could not explain the way an object glows depending on its temperature: its spectrum. · Max Planck explained it by assuming that atoms only emit radiation in quantum amounts...in steps given by the formula: E = hf where h is Planck’s constant and f is the frequency of the light h = 6.6 ´ 10 -34 J-s
Slide 16 / 83 Planck’s Quantum Hypothesis Planck didn't believe this was a real...it just worked. It was like working from the answers in the book...getting something that works, but having no idea why. It didn't make sense that atoms could only have steps of energy. Why couldn't they have any energy? Planck thought a "real" solution would eventually be found...but this one worked for some reason. Which brings us to our next mystery...
Slide 17 / 83 The Photoelectric Effect When light strikes a metal, electrons sometimes fly off. Classical physics couldn't explain some specific features about how the effect works. So Einstein used Planck's idea to solve it.
Slide 18 / 83 The Photoelectric Effect If atoms can only emit light in packets of specific sizes; maybe light itself travels as packets of energy given by Planck's formula. E = hf where h is Planck’s constant h = 6.6 ´ 10 -34 J-s or 4.14 x 10 -15 eV-s He called these tiny packets, or particles, of light, photons.
Slide 19 / 83 5 What is the energy (in nJ) of a photon with a frequency of 5 x 10 22 Hz?
Slide 20 / 83 The Photoelectric Effect The maximum kinetic energy of these photons can be measured using a variable voltage source and reversing the terminals so that the electrode C is negative and P is positive. If the voltage is increased, there is a point when the current reaches zero. This is called the stopping voltage, V 0 , and it is given by: KE max = eV 0
Slide 21 / 83 The Photoelectric Effect We said earlier that when light strikes a metal, electrons sometimes fly off. Since electrons are held in the metal by attractive forces, some minimum energy, W 0 , called the work function, is required just to get an electron free from the metal. The input energy of the photon will equal the kinetic energy of the ejected electron plus the energy required to free the electron. hf = KE + W 0
Slide 22 / 83 The Compton Effect A. H. Compton scattered short-wavelength light from various materials and discovered that the scattered light had a slightly lower frequency than the incident light, which indicated a loss of energy. He applied the laws of conservation of momentum and energy and found that the predicted results corresponded with the experimental results. Electron after A single photon wavelength, # , collision strikes an electron in some Incident material, knocking it out of its photon ( # ) # atom. The scattered photon # has less energy since it gave Electron at rest some to the electron and thus has a wavelength of , # '. Scattered photon ( #' )
Slide 23 / 83 The Compton Effect Electron The momentum of a photon is given by: after collision p = E/c Incident photon ( # ) # Since E = hf, # Electron at rest p = hf/c = h/ # Scattered photon ( #' ) Using conservation of momentum: Where m 0 is the rest mass of the electron.
Slide 24 / 83 Photon Interactions 1. The Compton Effect - the photon can be scattered by an electron and lose energy in the process. 2. The photoelectric effect - a photon can knock an electron out of an atom and disappear in the process. 3. The photon can knock and atomic electron to a higher energy state if the energy of the photon is not sufficient to knock it out of the atom. 4. Pair production - a photon can produce an electron and a positron and disappear in the process. (The inverse of pair production can occur if an electron collides with a positron. This is called annihilation.)
Slide 25 / 83 Photon Theory of Light This particle theory of light assumes that an electron absorbs a single photon and made specific predictions that proved true. For instance, the kinetic energy of escaping electrons vs. frequency of light shown below: This shows clear agreement with the photon theory, and not with wave theory. This shows that light is made of particles, photons; light is not a wave.
Slide 26 / 83 Wave-Particle Duality; the Principle of Complementarity Earlier we proved that light is a wave. Now we've proven that light is a particle. Which is it? This question has no answer; we must accept the dual wave-particle nature of light. While we cannot imagine something that is a wave and is a particle at the same time; that turns out to be the case for light.
Slide 27 / 83 6 The ratio of energy to frequency for a given photon gives A its amplitude. B its velocity. C Planck's constant. D its work function. E = hf c = l f h = 6.6 ´ 10 -34 J-s c = 3.00 ´ 10 8 m/s
Slide 28 / 83 7 What is a photon? A an electron in an excited state B a small packet of electromagnetic energy that has particle-like properties C one form of a nucleon, one of the particles that makes up the nucleus D an electron that has been made electrically neutral E = hf c = l f h = 6.6 ´ 10 -34 J-s c = 3.00 ´ 10 8 m/s
Slide 29 / 83 8 The energy of a photon depends on A its amplitude. B its velocity. C its frequency. D none of the given answers E = hf c = l f h = 6.6 ´ 10 -34 J-s c = 3.00 ´ 10 8 m/s
Slide 30 / 83 9 Which color of light has the lowest energy photons? A red B yellow C green D blue E = hf c = l f h = 6.6 ´ 10 -34 J-s c = 3.00 ´ 10 8 m/s
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