The Invention of the Modern Atom

1. The Invention of the Modern Atom

2. The First Atomic Theorist (Translation) “Everything is composed of ‘atoms’, which are physically, but not geometrically, indivisible. Between atoms lies empty space; atoms are indestructible; have always been, and always will be, in motion. There are an infinite number of atoms, and kinds of atoms, which differ in shape and size.” -Democritus, 350 BC V1

3. JJ Thomson (1897) Theorized that “cathode rays” are actually beams of subatomic particles. Thomson measured the mass of these particles by deflecting them in an electric field. It is now known that a cathode ray is a beam of electrons. V2

4. But Thomson noticed a large question brought up by his results. The knowledge at the time was the following. 1) It was known that electrons are negatively charged. 2) It was also known that atoms are (usually) net neutral. The question was: “Where does all the mass of an atom come from if the electrons have such small mass? How can the atom still be net neutral?” Thomson’s measured mass of an electron 9 × 10-31 kg Known mass of lightest element (Hydrogen) 1.7 × 10-27 kg

5. Thomson’s “Plum Pudding” Model An atom consists of negative electrons surrounded by a “pudding” of positive charge. The positively charged substance exactly balances out the charge of the electrons and is also massive enough to give the atom its measured mass.

6. Ernest Rutherford (1911) British particle physicist who fired alpha particles at a sheet of gold foil, and used fluorescent paper to detect where the alpha particles were deflected. V3

7. Rutherford’s Original Setup

8. Rutherford’s Surprising Setup Some of the alpha particles were being deflected at extreme angles, sometimes straight backward! With this one experiment, the Plum Pudding Model was finished.

9. Predicted by Thomson’s Plum Pudding Model Actual Results: Explained by Rutherford’s Planetary Model “The result of the experiment was as if you had fired a cannonball at a piece of tissue paper and had it come back and hit you.” –E.R.

10. Rutherford’s results (the large angles) are in line with the Planetary Model of the Atom, in which a dense, positively charged nucleus is surrounded by negative, low-mass electrons.

11. Big Problems with the Planetary Model When charged particles accelerate, they radiate electromagnetic waves – thus losing energy. If the electron were in circular motion (with centripetal acceleration), it would constantly radiate EM waves.

12. Spectrum Tube: The Fingerprint of an Element The tube is filled with a particular gas (Hydrogen, Helium, Neon, Mercury, etc) A voltage is applied across the tube, accelerating electrons through the gas. This causes some electrons to collide with electrons in gas atoms, transferring some energy to the atoms. This brings the gas atom into a higher energy state, otherwise known as an excited state. When the atom drops back to a lower energy, it emits a photon (which is what you see). S1

13. Spectroscope Time! By looking at the light through a diffraction grating, we can separate the spectrum to see which wavelengths are actually making up the observed color.

14. The Emission Spectrum of Hydrogen Hydrogen atoms, after gaining electric potential energy and and then emitting a photon as they return to a lower energy. We detect these frequencies of light during the process:

15. Niels Bohr “The Great Dane” Known for his dancing ability as well as his Nobel Prize in Physics.

16. What do you think Bohr came up with?

17. Bohr Model An electron can only occupy certainenergy states around a nucleus. When it transitions between energy states, a photon must either be absorbed or emitted by the atom. V4

18. Energy Level Diagram Each “level” represents a different electric potential energy for the electron/nucleus system. n = ∞ (ionization energy) 0 eV n = 2 (first excited state) n = 1 (first excited state) -19.36 eV -21.5 eV n = 0 (ground state) Caution! Do not take these diagrams too literally! These lines represent energy states for the atom.

19. Negative Potential Energies Again! Since an atom (nucleus and electrons) is in a bound state, it has a negative electric potential energy. This means that an input of energy is required to separate the electron and nucleus to d = ∞ (zero electric potential energy). The amount of energy that is needed to completely free an electron from the atom is called the ionization energy of the atom. (You are making the atom into a positive ion by freeing an electron) d ≈ ∞ Ground State (n = 1) Most negative EPE First Excited State (n = 2) Less negative EPE Ionization Energy (n = ∞) Zero EPE

20. The more negative the EPE, the closer the electron is (on average) from the nucleus. By putting energy into the system in the right amount, we can raise the system to a higher “energy state”. This will bring the electron (on average) further from the nucleus. n = ∞ (ionization energy) 0 eV n = 2 (first excited state) n = 1 (first excited state) -19.36 eV -21.5 eV n = 0 (ground state)

21. PhET Simulation – Hydrogen Atom

23. When an electron in an atom transitions from a more excited state (less negative energy) to a less excited state (more negative energy), it emits a photon!! n = 2 (first excited state) E2 = -3.4 eV The energy of the photon will be equal to the energy lost by the atom during this transition. E1 = -13.6 eV n = 1 (ground state)

24. Conservation of Energy works! Again! When an atom transitions to a lower energy level, it emits a photon. Einitial = Efinal + Ephoton Einitial Ephoton Efinal

25. Hydrogen Whiteboard Which color light is emitted during a transition from the n = 3 to the n = 2 energy level in hydrogen? n = 3 (second excited state) E3 = -1.5 eV E2 = -3.4 eV n = 2 (first excited state) E1 = -13.6 eV n = 1 (ground state)

26. Remember to convert to Joules! (h is given in terms of Joules!) n = 3 (second excited state) E3 = -1.5 eV E2 = -3.4 eV n = 2 (first excited state) (-1.5 eV) (1.6 x 10-19 J/eV) = (-3.4 eV) (1.6 x 10-19 J/eV) + hc/λ λ = 650 nm

27. Lots of different possibilities for energy level transitions!

28. For photons emitted during a drop in atomic EPE,

29. More Quantum Phenomena!! Now we have a new property that is quantized: the energy states of an atom. Atoms can only take on certain specific electric potential energies, and nothing in between. This is getting more interesting…

30. Predict the Spectrum of Neon! n = ∞ (ionization energy) n = 3 (first excited state) -16 eV n = 2 (first excited state) -19.36 eV What color will we see? What other wavelengths could we detect if we used IR and UV detectors? -21.5 eV n = 1 (ground state)

31. We can predict the wavelengths of the photons emitted during the transitions 3  2, 3  1, and 2 1. n = ∞ (ionization energy) n = 3 (first excited state) -16 eV n = 2 (first excited state) -19.36 eV -21.5 eV n = 1 (ground state)

32. PhET Simulation: Neon Lights

33. Lab Challenge II By watching the simulation, determine which energy transitions correspond with which photons (visible range only). Determine the changes in electric potential energy associated with the two visible photons emitted by excited sodium atoms. Fill in the numbers for the energy level diagram on your whiteboards!

34. Atomic Theory and the Bohr Model

36. Ultraviolet Visible Infrared

37. When an electron in an atom transitions from a more excited state (less negative energy) to a less excited state (more negative energy), it emits a photon!! n = 2 (first excited state) E2 = -3.4 eV The energy of the photon will be equal to the energy lost by the atom during this transition. E1 = -13.6 eV n = 1 (ground state)

38. Conservation of Energy works! Again! When an atom transitions to a lower energy level, it emits a photon. Einitial = Efinal + Ephoton Einitial Ephoton Efinal

39. Hydrogen Whiteboard Which color light is emitted during a transition from the n = 3 to the n = 2 energy level in hydrogen? n = 3 (second excited state) E3 = -1.5 eV E2 = -3.4 eV n = 2 (first excited state) E1 = -13.6 eV n = 1 (ground state) Visible Spectrum

40. Remember to convert to Joules! (h is given in terms of Joules!) n = 3 (second excited state) E3 = -1.5 eV E2 = -3.4 eV n = 2 (first excited state) (-1.5 eV) (1.6 x 10-19 J/eV) = (-3.4 eV) (1.6 x 10-19 J/eV) + hc/λ λ = 650 nm

41. Lots of different possibilities for energy level transitions!

42. For photons emitted during a drop in atomic EPE, Ef Ei λ

43. More Quantum Phenomena!! Now we have a new property that is quantized: the energy states of an atom. Atoms can only take on certain specific electric potential energies, and nothing in between. This is getting more interesting…

45. Predict the Spectrum of Neon! 0 eV n = ∞ (ionization energy) n = 3 (first excited state) -19.36 eV n = 2 (first excited state) -19.6 eV What color(s) will we see? What other wavelengths could we detect if we used IR and UV detectors? -21.5 eV n = 1 (ground state) Visible Spectrum

46. We can predict the wavelengths of the photons emitted during the transitions 3  2, 3  1, and 2 1. n = ∞ (ionization energy) n = 3 (first excited state) -19.36 eV n = 2 (first excited state) -19.6 eV -21.5 eV n = 1 (ground state)

47. What wavelengths can we expect to see? n = 2 to n = 1 First excited state to Ground state n = ∞ λ = 650 nm n = 3 -19.36 eV n = 2 -19.6 eV n = 3 to n = 1 Second excited state to Ground state λ = 580 nm n = 3 to n = 2 Second excited state to Ground state λ = 5,156 nm -21.5 eV n = 1 (Infrared)

48. Neon

49. Careful with Joules and eV! Use of Planck’s constant as h = 6.6 x 10-34 J*s and/or c = 3 x 108means that you cannot work in eV! You must then convert everything to J! Use this as your conversion factor: 1 J = 1.6 x 10-19 eV With quantum and atomic phenomena, you should always expect a reasonable number of eV for an answer (0-1000 eV). With quantum and atomic phenomena, you should always expect a very tiny number of Joules for an answer (5 x 10-19 J).

50. Ultraviolet Visible Infrared The larger the energy transition made by the electron, the more energy the emitted photon will have. This is why the “short” transitions of hydrogen (the Paschen series) emit infrared light. IR waves have long wavelengths and low-frequencies, thus carrying less energy. This is why the “long” transitions of hydrogen (the Lyman series) emit UV light. UV waves have short wavelengths and high-frequencies, thus carrying more energy. The visible portion spectrum (Balmer series) has energies right in the middle.