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CHAPTER 7

CHAPTER 7. QUANTUM THEORY OF THE ATOM. I. Rutherford's model of the atom. A) The nucleus is very small - positively charged - with the electrons outside the nucleus.

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CHAPTER 7

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  1. CHAPTER 7 QUANTUM THEORY OF THE ATOM

  2. I. Rutherford's model of the atom A) The nucleus is very small - positively charged - with the electrons outside the nucleus. B) A new question arises. If the electron is negatively charged, won't the attraction for electrons by the nucleus cause the electron to fall into the nucleus and therefore atoms should collapse.

  3. They don't. Why not? II. Rutherford attempts to explain his experimental results. A) He knew about the solar system - the attraction of the planets by the sun - universal gravitation.Yet planets are not pulled into the sun. They are in motion around the sun and this motion prevents them from being pulled into the sun. So Rutherford puts the electron in motion around the nucleus.

  4. B) There is a BIG difference between electrons and the nucleus and the sun and the planets. WHAT IS THAT??? C) An electron in an orbit experiences an acceleration. According to the laws of electricity, this moving negative charge in the vicinity of a positive charge will radiate energy - as it accelerates, it loses energy. What happens then?____________

  5. Problem for scientists since putting the electron in motion around the nucleus does not keep it outside the nucleus according to Newtonian Physics. D) The results of Rutherford's experiment cause a real dilemma for the scientists of the time who thought they had all the big problems solved.

  6. Only newer instruments were thought to be needed to get more precise measurements. III. CLUES TO THE SOLUTION OF THE PROBLEM A) Electromagnetic radiation 1) Sunlight is composed of visible electromagnetic radiation of wavelengths between 400 and 750 nm and other radiation as well. 2) Scientists had argued about the nature of light for years. Was it a______________ or a ____________?

  7. 3) At the beginning of the twentieth century, the wave theory predominated. This was the result of a diffraction experiment. Light waves add when they are in phase and subtract when out of phase. Matter was not known to subtract.

  8. 4) Characteristics of Waves  - wavelength - distance between consecutive peaks - crests - measured in m, nm, angstroms.

  9.  frequency - (nu) - number of times per second a crest passes a given point (cycles per second) 1 Hz = 1 cycle per second = 1/sec =sec-1 u = speed =  X  nm/wave X wave/sec = nm/sec for light - speed of electromagnetic radiation in a vacuum is a constant - c - 2.998 X 108 m/sec  X  = c for light

  10. nu is inversely proportional to the wavelength. What does this mean? The range of frequencies or wave lengths is called the electromagnetic spectrum - it ranges from gamma rays to TV, FM, AM radio waves.

  11. for visible light  is approximately 100 nm red 750 - 610 nm long  - low purple 450 - 400 nm short - high 

  12. 5) Visible light from an incandescent light bulb can be separated by a prism which bends light of different 's, different amounts. Shorter 's are bent more than longer 's. A continuous spectrum results from an incandescent light bulb.

  13. 6) Pass a current through a gas. The gas glows. When examined through a spectroscope, a line spectrum is observed, not a continuous one as seen with the light bulb.

  14. 7) Scientists named Angstrom and Balmer investigated these lines of hydrogen gas.

  15. 8) Through trial and error a formula was developed through which the wavelengths in the visible spectrum of hydrogen could be reproduced. 9) When attempts were made to apply this formula to the spectrum of atoms with more than one electron, it did not work.

  16. 10) The wave theory of light could not explain the line spectra of excited gases. Classical Newtonian Physics, which worked for moving automobiles, planets, etc. couldn't generate the Balmer Equation. B) Max Planck - explains black body radiation - i.e. radiation from heated solids. What solids in your home? kitchen? ???? Stars? The wavelength distribution depends on temperature.

  17. Planck tries to understand the relationship between intensity and wavelength of emitted light from hot objects. The then prevailing laws of physics could not account for what was observed. Planck made a daring assumption. Energy can be absorbed by atoms only in chunks of some minimum size. He gave the name quantum to the smallest quantity of energy that can be absorbed as electromagnetic radiation.

  18. He proposed that each chunk of energy carries a definite quantity: E = h The energy of one quantum is equal to Planck's constant times the speed of light divided by the wavelength. It must be set up to give Joules.

  19. According to Planck, energy is emitted in multiples of h. E = nh where n is an integer. This means that only whole numbers are allowed for n. So you can get 1 h, or 2 h, etc. , but never 0.5 h. or .2645 h. He cannot explain why energies associated with atoms should be fixed in this way. This is unusual. K.E. associated with cars, baseballs, trucks is continuous. In going from 0 to 45 miles per hour, you go through all speeds, not only certain ones.

  20. This is the stuff of science fiction where what happens to speeds?______________ C) Another mystery was solved by Einstein in 1905. 1) He explained the photoelectric effect. When light hits a metal, electrons are ejected.

  21. They are only ejected when the frequency of the light exceeds a certain threshold value for each particular metal. Violet light will cause potassium to eject electrons, no amount of red light, no matter how bright (intense) it is will have any effect. Einstein was surprised that the threshold energy was related to color rather than the intensity. More electrons are ejected with brighter light of a certain color, but the energy of each electron is the same.

  22. At the atomic level it appears that there is a 1 to 1 relationship between energy and phenomenon, not a continuum. 2) Einstein takes Planck's theory and extends it to the structure of light itself. He doesn't think of it as a continuous wave, but as chunks of 1wavelength. He calls these chunks of light, photons. Each photon has an energy E = to h

  23. IV. Bohr - 1913 - gave his model of the H atom – Nobel Prize awarded in 1922. A) A single electron moves in a circular orbit around the nucleus. To keep the e- from spiraling into the nucleus, Bohr says that only certain orbits are allowed which are at various distances from the nucleus.

  24. B) Electrons do not obey the laws of classical physics, therefore they do not lose energy in an orbit. C) In each orbit, the energy of the electron is restricted to a certain value - E = - RH/n2 where RH is a constant and n is an integer.

  25. D) Energy of the electron is quantized. Only certain energies are allowed to the electron. A staircase as opposed to an___________. E) Each orbit is characterized by an integer - 1, 2, 3, etc. RH is a constant in energy units: 2.179 X 10-18 J. F) When an electron changes orbits it changes energies.

  26. G) If energy is emitted: This is the energy of the emitted photon. Recall the Balmer-Rydberg equation, this equation now has a basis in Bohr's theory.

  27. H) Energy is emitted in the form of light (electromagnetic radiation as the electron moves from a higher orbit to a lower one (from a higher energy level to a lower one). I) Energy is absorbed as electricity or heat as the electron moves from a lower to a higher orbit (energy level). J) Energy levels are closer together as we move away from the nucleus. The difference between the energy levels becomes less and this is verified by the wavelengths observed.

  28. K) This leads to the birth of quantum mechanics (only certain energies allowed) – new theory for particles which barely exist. Remember the mass of the electron is about 1 X 10-28 g. V. PROBLEMS WITH THE BOHR ATOM A) It is only successful with the Hydrogen atom B) It could not account for extra lines in the H emission spectrum when a magnetic field was applied to the gas.

  29. C) The idea of circular orbits and knowing where the electron is located is impossible. VI. PARTICLE-WAVE DUALISM A) 1923-24 - The French physicist de Broglie says that if light waves exhibit particle properties, under certain circumstances, then particles of matter should show wave characteristics under certain circumstances.

  30. B) He postulated that a particle of mass m and speed v has a where h = 6.63 X 10-34 kg m2/sec if m is large and the speed is small, then  is so small as to be meaningless- 10-34 m.

  31. BUT, if mass is small, the mass of the electron for example 1 X 10-28 g, and the speed is large, then the  becomes measurable in picometers, 10-12 m. C) In 1927, Davisson and Germer in the U.S., at Bell Laboratories, and G.P.Thomson (son of J.J. who showed the electron was a particle) demonstrate that electrons have a wavelike nature. What did they have to do???

  32. Diffraction patterns were obtained by reflecting electrons off crystals as well as passing electrons through thin gold foil. These patterns were similar to those obtained when passing X-rays through crystals.

  33. It is interesting to note that J.J. Thomson received a Nobel Prize for the showing that the electron was a particle and his son G.P. Thomson received a Nobel Prize for showing that the electron can be considered a wave. Schrödinger - 1926 -awarded Nobel Prize in 1933 A) He developed a more powerful model of the atom.

  34. B) He combined the equations for the behavior of waves with the de Broglie equation to generate a mathematical model for the distribution of electrons in atoms. C) The advantage of Schrödinger's approach is that it consists of mathematical equations known as wave functions that satisfy the requirements placed on the behavior of electrons. D) The disadvantage is that it is difficult to imagine electrons at waves.

  35. E) Schrödinger's equation provides information about an electron's location in terms of probability of finding an electron which has a certain energy in a given location. Why not certainty? F) In 1927 Werner Heisenberg showed from quantum mechanics that it is impossible to know simultaneously, with absolute precision, both the position and the momentum of a particle such as an electron. The Heisenberg Uncertainty Principle!

  36. G) The Bohr model was a 1 dimensional model (the radius) that used 1 quantum number to describe the distribution of electrons in an atom. The only important information was the size of the orbit which was described by the quantum number n. H) Schrödinger's model allows the electron to occupy 3D space. It requires 3 coordinates or 3 quantum numbers to the describe the ORBITALS in which the electrons can be found..

  37. I) The three coordinates that come from the wave equation of Schrödinger are the principal quantum number (n),the angular quantum number ( ),and the magnetic quantum number (m). A fourth quantum number arises out of relativistic effects (Einstein's Theory of Relativity),the spin quantum number,(ms). J) The 4 quantum numbers are like a zip code for the electron.

  38. K) They specify an atomic orbital, a region in space where there is high probability of finding an electron with a characteristic energy, and the number of electrons which can occupy the orbital. VIII. THE QUANTUM NUMBERS A) The Principal Quantum Number - the modern equivalent of n in the Bohr Theory. It describes the main energy level. It can have the values of the positive integers: 1, 2, 3, 4, 5,....

  39. It is related to the average distance of the electron from the nucleus. The energy of the electron depends principally on n. Orbitals of the same quantum number n, belong to the same shell. B) Angular momentum quantum number  - azimuthal or subsidiary quantum number - distinguishes orbitals of a given n having different shapes. Other synonyms are sublevel and subshell. There are n different kinds of orbitals each with a distinctive shape denoted by  .

  40.  has values from 0 to n-1. (It is important to remember that in this case 0 does not mean nothing.) When n = 1,  can only equal 0 - only one subshell When n = 2,  can equal 0 and 1 - two subshells When n = 3,  can equal 0, 1, and 2 - three subshells

  41. Associated with each value of  is a letter related to a shape which is a region of space with an approximate 90% occupancy rate by an electron of a specified energy. When = 0, the letter designation is s and the shape is spherical. When = 1, the letter designation is p and the shape is dumbbell shaped. When  = 2, the letter designation is d and the shape is a cloverleaf and another shape.

  42. designations used are 1s, 2s, 2p, 3s, 3p, 3d,...

  43. C) m is the magnetic quantum number which distinguishes orbitals of given n and . It specifies the orientation in space of the atomic orbital. The number of different orientations in space depends on the subshell designated. The allowed values are integers from -  through 0 to +  giving 2  +1 possibilities. When n = 1;  = 0; m  = 0 - only 1 orientation possible - a sphere.

  44. When n = 2;  = 0; m = 0 - only 1 orientation possible - a sphere. When n = 2;  = 1; m  = -1, 0, +1 - 3 orientations are possible – one dumbbell along each of the three axes, x, y and z.

  45. When n = 3; = 0; m  = 0 - only 1 orientation possible - a sphere. When n = 3; = 1; m  = -1, 0, +1 The 3 orientations which are possible are one dumbbell along each of the three axes, x, y, z. When n = 3; = 2; m = -2 -1, 0, +1 +2 There are 5 orientations possible – four cloverleafs and 1 other shape. One is along the xy axes, three between the axes, and the special one is along the z axis.

  46. D) ms is the spin quantum number. An electron has magnetic properties that correspond to a charged particle spinning in its axis. Either of 2 spins are possible??? 2 values are possible - +½ and -½ for every set of n,  , m - this gives two as the number of electrons which can occupy each orbital. 2 e's in the 1s orbital - "zip code" 1,0,0,+ ½ and 1,0,0,- ½ .

  47. 2 e's in the 2s orbital - "zip code" 2,0,0,+ ½ and 2,0,0,- ½ . 2 e's in each 2p orbital - "zip code" 2,1,-1,+½; 2,1,-1,- ½; 2,1,0,+ ½; 2,1,0, -½; 2,1,+1,+ ½; 2,1,+1- ½ .

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