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Quantum Mechanical Ideas

Quantum Mechanical Ideas. Photons and their energy. When electromagnetic waves are exhibiting their “particle-like” nature, we call those little mass-less bundles of energy PHOTONS. There are photons of light, photons of UV, photons of microwaves, photons of IR, etc. New Units of measurement.

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Quantum Mechanical Ideas

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  1. Quantum Mechanical Ideas

  2. Photons and their energy When electromagnetic waves are exhibiting their “particle-like” nature, we call those little mass-less bundles of energy PHOTONS. There are photons of light, photons of UV, photons of microwaves, photons of IR, etc.

  3. New Units of measurement Sometimes the wavelengths of photons are measured in meters, sometimes in nanometers and sometimes in Angstroms, where one Angstrom = 1 x 10-10 meters Also, the ENERGY of electrons is often given in “electron-Volts”, eV, instead of Joules, where one eV = 1.6 x 10-19 J- a very tiny amount of energy!

  4. All electromagnetic photons carry energy as they travel along at “the speed of light”. The energy of a photon, in eV, is given by E = hf where f is the frequency of the photon, measured in Hertz h is a constant called Plank’s constant. h = 4.14 x 10-15 eV·s Or h = 6.63 x 10-34 J·s

  5. For photons,like “visible light”, UV, IR, microwaves, etc.c = lf where c = the “speed of light” c = 3 x 108 m/s

  6. The difference frequency of electromagnetic waves (photons) determines if they are visible light, radio wave, microwaves, etc. higher frequency = more energy!

  7. Which photon has more energy- an X-ray photon or a microwave photon?

  8. The different frequencies of visible light correspond to different colors of light. Blue light has a higher frequency than yellow light. Which color of light has the highest energy?

  9. How can you produce different colors of light?

  10. What makes one atom different from another? The amazing colors produced in fireworks are a result of the different types of atoms that are used to make the fireworks.

  11. Each atom has its own unique number of protons, neutrons, and electrons. Each electron in every element is in an “orbital” about the nucleus and has a unique energy. That unique energy determines the amazing colors seen in fireworks!

  12. A Quantum is a discreet unit of a physical quantity. • For example: our money is measured in a quantum of one cent. You can have 1 cent, 2 cents, 8 cents, etc., but you can’t have 1.24 cents or 19.68 cents! You must jump from 1 to 2 to 3 to 4, etc. • Electric charge, which ultimately comes from either a proton or an electron, is QUANTIZED. • There is no such thing as a half of an electron or a fifth of a proton, so everything that has electrical charge must have some multiple of the charge of an electron or proton- 5 electrons, 8 protons, etc. That’s why electric charge is QUANTIZED.

  13. The electrons, in their orbitals about the nucleus, have QUANTIZED levels of energy that are determined by which orbital they are in. The orbitals are numbered with “n” numbers, the “principle quantum number”: n = 1, n = 2, n = 3, etc. where the orbital closest to the nucleus is n = 1. The “n-number” for each atom’s electrons determine that electron’s energy. The larger the “n”, the larger the energy.

  14. What does the energy of an electron in its orbital have to do with the colors of fireworks???

  15. When an electron absorbs energy from an external source in any form (heat, electricity, a collision, etc.), it jumps to a higher orbital- called an “excited state”.

  16. When the electron falls back down to its original orbital, called its “rest state” or “ground state”, it must give up that extra energy. The energy is emitted in the form of a photon! Some of those emitted photons are visible light of different colors- some photons are not visible to us, like UV or IR or microwaves or X-rays

  17. If an atom is continually absorbing energy, all kinds of transitions between higher and lower orbital levels are possible, resulting in many different types of emitted photons of many different colors.

  18. What elements are used in fireworks to produce different colors of light?

  19. Atomic Spectra • Since the electrons’ energy are unique for each element, each element produces a unique spectra of colors when supplied energy. • We may see with our eyes only many overlaping colors of light. To see all the distinct colors in the atom’s spectra requires a “diffraction grating”. Spectra for Neon

  20. Each element produces a unique spectra of colored lines when viewed through a diffraction grating Argon Nitrogen Mercury Helium

  21. Because each element produces a unique emission spectra, scientists use “spectral analysis” to determine the composition of unknown substances. The spectra is like a fingerprint- absolutely unique for each element! Astronomers use “spectral analysis” to determine the composition of stars as well. Argon

  22. Using a Spectrometer to determine the identity of a elemental gas • The gas will not glow until it is energized. Energy can be provided in the form of heat or by applying a high voltage. The spectrum analysis power supply shown here provides high voltage. • If you look at the glowing tube with just a diffraction grating, the emission spectrum lines of color are visible. • If you look at the glowing tube through a “spectrometer”, which contains a diffraction grating, you can actually precisely measure the angles between the lines. • Those angles allow you to precisely determine the wavelengths (or frequencies) of each of those colors. • Since each element emits only certain wavelengths, the gas can be identified.

  23. Hydrogen The emission spectrum of Hydrogen is the most studied spectrum because it is also the simplest. Hydrogen has only ONE electron. But that ONE electron can be energized to many different orbitals, “excited states” and will emit photons as it returns to its “ground state”.

  24. Suppose an electron makes a transition from n = 3 to n = 2. What is the energy of the emitted photon? Energy = E3 – E2 E = 12.07 eV – 10.19 eV E = 1.88 eV What is the energy of an emitted photon if an electron makes a transition from n = 4 to n = 1? E = 12.73 eV The higher the energy of the photon, the higher its frequency! E = hf

  25. Each color has a different energy. The further apart the lines, the greater the difference in energy. The closer the lines, the less the difference in energy. Look at the spectrum for Hydrogen. Which two lines have the least difference in energy? Which two lines have the greatest difference in energy?

  26. Atomic Spectra Absorption of an external source of energy results in a transition to a higher energy level. • A transition back to a lower level must release energy – in the form of a photon! • The frequency of the emitted photon is determined by the difference in the energy levels. Ephoton = E2 – E1 Since E = hf, the higher the energy, the higher the frequency! • Different frequencies are different colors of light or different types of EM Waves

  27. The Hydrogen Atom:ONEelectron! The frequency of the emitted photon is determined by the difference in the energy levels. Ephoton = hf = E2 – E1

  28. Now you try one…. • Get out your calculators!! • The energized electron in Hydrogen makes a transition from n = 3 with an energy of -1.5 eV down to its ground state where its energy is -13.6 eV. What is the frequency of the emitted photon? • Ephoton = E2 – E1 and Ephoton = hf • Ephoton = -1.5 eV – (-13.6 eV) = 12.1 eV • Ephoton = 12.1 eV = hf (h = 4.14 x 10-15 eV·s ) • f = 12.1 eV ÷ 4.14 x 10-15 eV·s = • Frequency, f = 2.92 x 1015 Hz • Is this visible light?? Use c = lf to find the wavelength. • wavelength, l = 1.02 x 10-7 = 102 nm • This is NOT visible light- it is UV.

  29. Hydrogen Emission The energy, in eV, of the electron, in a Hydrogen atom is given by E = 1.Find the energy of the electron at each orbit from n = 1 to n = 7. 2. Find the Energy of eachphoton emitted (Ephoton = DE) for these transitions from one orbit to another: 4 to 1 5 to 2 6 to 3 7 to 4 3 to 1 4 to 2 5 to 3 6 to 4 2 to 1 3 to 2 4 to 3 5 to 4 3. Find the wavelength for each of those photons and determine what type of electromagnetic wave they are.

  30. Light behaves like a wave AND like a particle The first clear demonstration of the particle-like behavior of light was in The Photoelectric Effect Albert Einstein won the Nobel Prize in Physics for his study of the Photoelectric Effect.

  31. Shining light on a metal can liberate electrons from its surface. The light has to have enough energy (high enough frequency) for this effect to occur. The energy of the “photoelectrons” liberated from the surface depends on the frequency (the energy) of that incident light- NOT its intensity! Increasing the intensity of the light increases the number of photoelectrons emitted, but not the energy of each photoelectron.

  32. When will Photoelectrons be produced? PHet simulation (Go to PHet website to explore the photoelectric effect simulation)

  33. If no electrons are ejected, you must… …increase the frequency of the light If only a few electrons are ejected and you want more, your must….. …increase the intensity of the light If you want to increase the kinetic energy of the electrons, you must… …increase the frequency of the light.

  34. … and the math… Using conservation of energy: The energy of the incident photon disappears. Where does it go? First, that energy must be used to liberate the electron. That energy is called the WORK FUNCTION, WO. Each kind of metal has its unique work function. Any extra energy is given appears in the electron as kinetic energy, K Photon Energy = Work function + Kinetic energy E = Wo + K

  35. ExamplePhoton energy = Wo + K A photon with energy 3.2 eV strikes a metal surface with a work function of 1.8 eV. What is the kinetic energy of the ejected photoelectrons? K = photon energy – Wo K = 3.2 eV – 1.8 eV K = 1.4 eV

  36. ExamplePhoton energy = Wo + K A photon with energy 2.8 eV strikes a metal surface. If the kinetic energy of the ejected photoelectrons is 0.5 eV, what is the work function of the metal? Wo = photon energy - K Wo = 2.8 eV – 0.5 eV Wo = 2.3 eV

  37. hf = Wo + ½ mv2 There is a minimum frequency, called the “threshold frequency” required to liberate an electron. At the threshold frequency, Energy of photon = hfthreshold = Wo That threshold photon’s wavelength is called the “cutoff wavelength” and can be found using c = lf

  38. Now you try some: If the cut-off wavelength for a particular metal is 320 nm, what is the metal’s work function? First, find the threshold frequency using c = lf Threshold frequency, fo = 9.38 x 1014 Hz Now find the work-function: hfo = Wo Be careful to use Planck’s constant with the correct units: h = 6.63 x 10-34J·s or h = 4.14 x 10-15 eV·s Wo = 6.22 x 10-19 J or Wo = 3.88 eV

  39. What if a 450 nm light hit a surface with a work function of 2.36 eV. What will be the kinetic energy of the photoelectron? First, find the frequency of the 450 nm light. f = 6.67 x 1014 Hz Now, using conservation of energy: hf = Wo + Kinetic energy So K = hf – Wo K = 0.40 eV How fast is the ejected electron moving? K = ½ mv2 m = 9.1 x 10-31 kg AND you have to convert K back into Joules first!

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