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Modern Physics

Modern Physics. What is meant by “ quantized ”? Quantity Specific and discrete quantity Packets of definite size. Quantized energy can be thought of as existing in very small packets of specific size. Atoms absorb and emit quanta of energy. Let us consider light …

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Modern Physics

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  1. Modern Physics

  2. What is meant by “quantized”? • Quantity • Specific and discrete quantity • Packets of definite size. • Quantized energy can be thought of as existing in very small packets of specific size. • Atoms absorb and emit quanta of energy.

  3. Let us consider light… Electromagnetic Spectrum Visible spectrum

  4. Three models are used to describe light • What model is used in geometric optics, like with lenses and mirrors? • Ray • What model is used in studying diffraction and interference? • Wave • What model is used to study interaction of light and atoms? • Particle ( photon)

  5. But light is said to have a “dual nature” • What is that supposed to mean? • Wave particle duality • Waves have both a wave and a particle component • We describe the path of light as a ray • Equation • E = hf for a single photon • E = nhf for multiple photons • h = plank’s constant • 6.63x10-34 J ∙ s (SI version) • 4.14x10-15eV ∙ s (convenient) • f = frequency • n = number of photons

  6. Conceptual checkpoint • Which has more energy in its photons, a very bright, powerful red laser or a small key-ring red laser? • Neither! They both have the same energy per photon. The big one has more power. • Which has more energy in its photons, a red laser or a green laser? • The green one has shorter wavelength and higher frequency. It has more energy per photon.

  7. The “electron-volt” (eV) • The electron-volt is the most useful unit on the atomic level. • If a moving electron is stopped by 1 V of electric potential, we say it has 1 electron-volt (or 1 eV) of kinetic energy. • 1 eV = 1.602×10-19 J

  8. E = 4.0x10-19 J h= 6.625x10-34 J ∙ s • What is the frequency and wavelength of a photon whose energy is 4.0 x 10-19J? E = hf f = E / h = 4.0x10-19 J / 6.625x10-34 J ∙ s = 6.04x1014 Hz λ = c/f λ = c/f = 3x108 m/s / 6.04x1014 1 / s = 4.97x10-7 m = 497 nm

  9. P = 3.0 mW = 0.003 W λ = 632.8 nm = 632.8x10-9 m • How many photons are emitted per second by a He-Ne laser that emits 3.0 mW of power at a wavelength of 632.8 nm? P = E /t E = P ∙ t E = 0.003 W ∙ 1 s = 0.003 J f = c / λ = 3x108 m/s / 632.8x10-9 m = 4.74x1014 Hz E =n(hf) n =E / (hf) =0.0003 J / (6.625x10-34 J ∙ s ∙ 4.74x1014 Hz) = 9.55x1014

  10. What are atoms composed of? • Atoms consist of nuclei (protons and neutrons and electrons. • What happens when an atom encounters a photon? • The atom usually ignores the photon, but sometimes the atom absorbs the photon. • If the photon is absorbed by the atom, what happens next? • The photon disappears and winds up giving all its energy to the atom’s electrons.

  11. Atom loses an electron Ionization level 0.0eV Third excited state -1.0eV • This is a graph of energy levels for a hypothetical atom Highest energy level Second excited state -3.0eV Higher energy levels. (Atoms has to absorb energy to get from the ground state. First excited state -5.5 eV Normal “unexcited” state Ground state (lowest energy level -11.5 eV

  12. Ionization level 0.0eV • Only certain energies are allowed. • These are represented by the horizontal lines. • The atom cannot exist at energies not shown in this graph! Third excited state -1.0eV • What do we mean when we say the atoms energy levels are “quantized”? Second excited state -3.0eV First excited state -5.5 eV Ground state (lowest energy level -11.5 eV

  13. Absorption of photon by Atom • When a photon of light is absorbed by an atom, it causes an increase in the energy of the atom. • The photon disappears, and the energy of the atom increases by exactly the amount of energy contained in the photon. • The photon can be absorbed ONLY if it can produce an “allowed” energy increase in the atom.

  14. 0.0eV Exited state • Absorption of photon by atom ΔE = hf E = hf Ground state -10.0eV

  15. Absorption Spectrum • When an atom absorbs photons, it removes the photons from the white light striking the atom, resulting in dark bands in the spectrum. • Therefore, a spectrum with dark bands in it is called an absorption spectrum.

  16. Ionized 0.0eV • Absorption Spectrum Absorption spectra always involve atoms going up in energy level. Ground state -10.0eV

  17. Emission of photon by atom • When a photon of light is emitted by an atom, it causes a decrease in the energy of the atom. • A photon of light is created, and the energy of the atom decreases by exactly the amount of energy contained in the photon that is emitted. • The photon can be emitted ONLY if it can produce an “allowed” energy decrease in an excited atom.

  18. 0.0eV Exited state • Emission of photon by atom ΔE = hf E = hf Ground state -10.0eV

  19. Emission Spectrum • When an atom emits photons, it glows! The photons cause bright lines of light in a spectrum. • Therefore, a spectrum with bright bands in it is called an emission spectrum.

  20. Ionized 0.0eV • Emission of photon by atom Emission spectra always involve atoms going down in energy level. Ground state -10.0eV

  21. Photoelectric effect

  22. What is the frequency and wavelength of the light that will cause the atom shown to transition from the ground state to the first excited state? Draw the transition. Ionization level h = 4.14x10-15eV ∙ s 0.0eV ΔE = hf f = Δ E / h Third excited state -1.0eV f = (11.5 – 3.0) / 4.14x10-15 Second excited state -3.0eV f = 1.45x1015 Hz First excited state -5.5 eV λ= c / f λ= 3x108 / 1.45x1015 Ground state (lowest energy level λ= 2.07x10-7 m λ= 207 nm -11.5 eV

  23. What is the longest wavelength of light that when absorbed will cause the atom shown to ionize from the ground state? Draw the transition. Ionization level h = 4.14x10-15eV ∙ s 0.0eV ΔE = hf f = Δ E / h Third excited state -1.0eV f = (11.5 ) / 4.14x10-15 Second excited state -3.0eV f = 2.78x1015 Hz First excited state -5.5 eV λ= c / f λ= 3x108 / 2.78x1015 Ground state (lowest energy level λ= 1.08x10-7 m λ= 1-8 nm -11.5 eV

  24. The atom shown is in the second excited state . What frequencies of light are seen in its emission spectrum? Draw the transition. Ionization level h = 4.14x10-15eV ∙ s 0.0eV Third excited state 1 -1.0eV ΔE = hf f = Δ E / h Second excited state -3.0eV f = (11.5 – 3.0 ) / 4.14x10-15 1 2 f = 2.053x1015 Hz First excited state -5.5 eV 3 Ground state (lowest energy level -11.5 eV

  25. The atom shown is in the second excited state . What frequencies of light are seen in its emission spectrum? Draw the transition. Ionization level h = 4.14x10-15eV ∙ s 0.0eV Third excited state 2 -1.0eV ΔE = hf f = Δ E / h Second excited state -3.0eV f = (5.5-3 ) / 4.14x10-15 1 2 f = 6.09x1014 Hz First excited state -5.5 eV 3 Ground state (lowest energy level -11.5 eV

  26. The atom shown is in the second excited state . What frequencies of light are seen in its emission spectrum? Draw the transition. Ionization level h = 4.14x10-15eV ∙ s 0.0eV Third excited state 3 -1.0eV ΔE = hf f = Δ E / h Second excited state -3.0eV f = (11.5-5.5 ) / 4.14x10-15 1 2 f = 1.45x1015 Hz First excited state -5.5 eV 3 Ground state (lowest energy level -11.5 eV

  27. 1 f = 2.053x1015 Hz The atom shown is in the second excited state . What frequencies of light are seen in its emission spectrum? Draw the transition. Ionization level 0.0eV Third excited state -1.0eV 2 f = 6.09x1014 Hz Second excited state -3.0eV 1 2 First excited state 3 f =1.45x1015 Hz -5.5 eV 3 Ground state (lowest energy level -11.5 eV

  28. We’ve seen that if you shine light on atoms, they can absorb photons and increase in energy. • The transition shown is the absorption of an 8.0 eV photon by this atom. • You can use Planck’s equation to calculate the frequency and wavelength of this photon. Atoms absorbing photons increase in energy Ionization level 0.0eV Photon - 4.0 eV with largest allowed energy -4.0eV Ground state (lowest energy level -12 eV

  29. Question • Now, suppose a photon with TOO MUCH ENERGY encounters an atom? • If the atom is “photo-active”, a very interesting and useful phenomenon can occur… • This phenomenon is called the Photoelectric Effect.

  30. Some “photoactive” metals can absorb photons that not only ionize the metal, but give the electron enough kinetic energy to escape from the atom and travel away from it. • The electrons that escape are often called “photoelectrons”. • The binding energy or “work function” is the energy necessary to promote the electron to the ionization level. • The kinetic energy of the electron is the extra energy provided by the photon. E = W0 + KE e- Eph Kinetic energy Photoelectric Effect 0.0eV Ionization level -4.0eV W0 = Work function Photon energy Ground state (lowest energy level -12 eV

  31. Photon Energy = Work Function + Kinetic Energy • hf = Φ + Kmax • Kmax = hf – Φ • K:max Kinetic energy of “photoelectrons” • hf: energy of the photon • Φ: binding energy or “work function” of the metal. E = W0 + KE e- Eph Kinetic energy Photoelectric Effect 0.0eV Ionization level -4.0eV W0 = Work function Photon energy Ground state (lowest energy level -12 eV

  32. Suppose the maximum wavelength a photon can have and still eject an electron from a metal is 340 nm. What is the work function of the metal surface? The longest wavelength is the lowest energy, and will provide no “extra” kinetic energy for the electron. Kmax = hf – Φ 0 J = hf – Φ f = v / λ Φ = hv / λ Φ = hf Φ = (4.14x10-15eV ∙ 3x108 m/s) / 340x10-9 m Φ = 3.65 eV

  33. Question • Suppose you collect Kmaxand frequency data for a metal at several different frequencies. You then graph Kmax for photoelectrons on y-axis and frequency on x-axis. What information can you get from the slope and intercept of your data? Slope: Planck’s Constant Intercept: Φ - binding energy or “work function”

  34. The Photoelectric Effect experiment • The Photoelectric Effect experiment is one of the most famous experiments in modern physics. • The experiment is based on measuring the frequencies of light shining on a metal (which is controlled by the scientist), and measuring the resulting energy of the photoelectrons produced by seeing how much voltage is needed to stop them. • Albert Einstein won the Nobel Prize by explaining the results.

  35. Light Collector Photoelectric Effect experiment Metal (+) e- e- e- e- e- e- e- e- e- e- e- V e- e- e- e- e- e- e- e- e- A e- e-

  36. Strange results in the Photoelectric Effect experiment • Voltage necessary to stop electrons is independent of intensity (brightness) of light. It depends only on the light’s frequency (or color). • Photoelectrons are not released below a certain frequency, regardless of intensity of light. • The release of photoelectrons is instantaneous, even in very feeble light, provided the frequency is above the cutoff.

  37. Number of electrons (current) increases with brightness, but energy of electrons doesn’t! Voltage versus current for different intensities of light. “Stopping Voltage” Vs, the voltage needed to stop the electrons, doesn’t change with light intensity. That means the kinetic energy of the electrons is ndependent of how bright the light is.

  38. Energy of electrons increases as the energy of the light increases. Voltage versus current for different frequencies of light. “Stopping Voltage” Vs changes with light frequency. That means the kinetic energy of the photoelectrons is dependent on light color.

  39. Experimental determination of the Kinetic Energy of a photoelectron • The kinetic energy of photoelectrons can be determined from the voltage (stopping potential) necessary to stop the electron. • If it takes 6.5 Volts to stop the electron, it has 6.5 eV of kinetic energy.

  40. Kmax Graph of Photoelectric Equation KMAX = hf - Φ y= mx + b Slope = h f Cut-off Frequency Φ (binding energy)

  41. Question: Does a photon have mass? • A photon has a fixed amount of energy (E = hf). • We can calculate how much mass would have to be destroyed to create a photon (E=mc2).

  42. λ = 340x10-9 m h = 6.63x10-34 J∙s • Calculate the mass that must be destroyed to create a photon of 340 nm light. h = 6.63x10-34 (kg ∙m2 / s2)∙s E = hf f = c / λ E = mc2 hf = mc2 hc / λ = mc2 h / λ = mc m = h / (λc) m = 6.63x10-34 / (340x10-9 ∙3x108 ) =

  43. Momentum of a Photon • Does a photon have momentum? Yes • A photon’s momentum is calculated by • p = E / c = hf / c = h / λ

  44. We have experimental proof of the momentum of photons • Compton scattering • Proof that photons have momentum. • High-energy photons collided with electrons exhibit conservation of momentum. • Work Compton problems just like other conservation of momentum problems – except the momentum of a photon uses a different equation.

  45. melectron = 9.11x10-31 Kg h = 6.63x10-34 J∙s • What is the frequency of a photon that has the same momentum as an electron with speed 1200 m/s? h = 6.63x10-34 (kg ∙m2 / s2)∙s • p = mv • p = 9.11x10-31 Kg ∙ 12 m / s • = • p = hf / c • f = P c / h • f = ( kg ∙ m/s) ∙ (3x108 m/s) / (6.63x10-34 (kg ∙m2 / s2)∙s) • f =

  46. Wave-Particle Duality • Waves act like particles sometimes and particles act like waves sometimes. • This is most easily observed for very energetic photons (gamma or x-Ray) or very tiny particles (elections or nucleons)

  47. Particles and Photons both have Energy • A moving particle has kinetic energy • E = K = ½ mv2 • A particle has most of its energy locked up in its mass. • E = mc2 • A photon’s energy is calculated using its frequency • E = hf

  48. Particles and Photons both have Momentum • For a particle that is moving • p = mv • kg ∙ m/s • For a photon • p = h/λ • (kg ∙m2 / s2)∙s / m = kg ∙ m / s • Check out the units! They are those of momentum.

  49. Particles and Photons both have a Wavelength • For a photon • l = c/f • For a particle, which has an actual mass, this equation still works • λ= h/p where p = mv • This is referred to as the deBroglie wavelength

  50. We have experimental proof that particles have a wavelength • Davisson-Germer Experiment • Verified that electrons have wave properties by proving that they diffract. • Electrons were “shone” on a nickel surface and acted like light by diffraction and interference. • We’ll study diffraction in the next unit, and return to this experiment then…

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