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The Invention of the Modern Atom 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 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 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 Known mass of lightest element (Hydrogen) 9 × 10-31 kg 1.7 × 10-27 kg 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. 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 Rutherford’s Original Setup 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. 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. 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. 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. 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 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. 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: Niels Bohr “The Great Dane” Known for his dancing ability as well as his Nobel Prize in Physics. What do you think Bohr came up with? Bohr Model An electron can only occupy certain energy states around a nucleus. When it transitions between energy states, a photon must either be absorbed or emitted by the atom. V4 Energy Level Diagram Each “level” represents a different electric potential energy for the electron/nucleus system. 0 eV n = ∞ (ionization energy) n = 2 (first excited state) -19.36 eV -21.5 eV n = 1 (first excited state) n = 0 (ground state) Caution! Do not take these diagrams too literally! These lines represent energy states for the atom. 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 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. 0 eV n = ∞ (ionization energy) n = 2 (first excited state) -19.36 eV -21.5 eV n = 1 (first excited state) n = 0 (ground state) PhET Simulation – Hydrogen Atom 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!! E2 = -3.4 eV n=2 (first excited state) 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) Conservation of Energy works! Again! When an atom transitions to a lower energy level, it emits a photon. Einitial = Efinal + Ephoton Einitial Einitial = E final + hc Ephoton l Efinal Hydrogen Whiteboard Which color light is emitted during a transition from the n = 3 to the n = 2 energy level in hydrogen? E3 = -1.5 eV n=3 (second excited state) E2 = -3.4 eV n=2 (first excited state) E1 = -13.6 eV n=1 (ground state) Einitial = E final + hc l Remember to convert to Joules! (h is given in terms of Joules!) E3 = -1.5 eV n=3 (second excited state) 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 Lots of different possibilities for energy level transitions! For photons emitted during a drop in atomic EPE, Ei = E f + lemitted hc lemitted hc = Ei - E f 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… Predict the Spectrum of Neon! n = ∞ (ionization energy) -16 eV n = 3 (first excited state) -19.36 eV n = 2 (first excited state) 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) We can predict the wavelengths of the photons emitted during the transitions 3 2, 3 1, and 2 1. n = ∞ (ionization energy) -16 eV n = 3 (first excited state) -19.36 eV n = 2 (first excited state) Ei = E f + -21.5 eV n = 1 (ground state) hc lemitted PhET Simulation: Neon Lights 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! Atomic Theory and the Bohr Model Ultraviolet Visible Infrared 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!! E2 = -3.4 eV n=2 (first excited state) 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) Conservation of Energy works! Again! When an atom transitions to a lower energy level, it emits a photon. Einitial = Efinal + Ephoton Einitial Einitial = E final + hc Ephoton l Efinal Hydrogen Whiteboard Which color light is emitted during a transition from the n = 3 to the n = 2 energy level in hydrogen? E3 = -1.5 eV n=3 (second excited state) E2 = -3.4 eV n=2 (first excited state) E1 = -13.6 eV Visible Spectrum n=1 (ground state) Einitial = E final + hc l Remember to convert to Joules! (h is given in terms of Joules!) E3 = -1.5 eV n=3 (second excited state) 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 Lots of different possibilities for energy level transitions! For photons emitted during a drop in atomic EPE, Ei = E f + lemitted hc lemitted hc = Ei - E f Ef Ei λ 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… Predict the Spectrum of Neon! 0 eV n = ∞ (ionization energy) -19.36 eV n = 3 (first excited state) -19.6 eV n = 2 (first excited state) 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 We can predict the wavelengths of the photons emitted during the transitions 3 2, 3 1, and 2 1. n = ∞ (ionization energy) -19.36 eV n = 3 (first excited state) -19.6 eV n = 2 (first excited state) Ei = E f + -21.5 eV n = 1 (ground state) hc lemitted lemitted hc = Ei - E f What wavelengths can we expect to see? n=∞ -19.36 eV n=3 -19.6 eV n=2 n = 2 to n = 1 First excited state to Ground state λ = 650 nm 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 -21.5 eV n=1 λ = 5,156 nm (Infrared) Neon Careful with Joules and eV! Use of Planck’s constant as h = 6.6 x 10-34 J*s and/or c = 3 x 108 means 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). 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 lowfrequencies, 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. A hypothetical atom has four energy states as shown below. 1)Which of the following photon energies could NOT be found in the emission spectra of this atom after it has been excited to the n = 4 state? (A) 1 eV (B) 2 eV (C) 3 eV (D) 4 eV (E) 5 eV 2) Which of the following transitions will produce the photon with the longest wavelength? (A) n = 2 to n = 1 (B) n = 3 to n = 1 (C) n = 3 to n = 2 (D) n = 4 to n = 1 (E) n = 4 to n = 3 Once the atom has been excited to n = 4, all of these possible transitions can occur! Which of the following photon energies could NOT be found in the emission spectra of this atom after it has been excited to the n = 4 state? (A) 1 eV (B) 2 eV (C) 3 eV (D) 4 eV (E) 5 eV None of these possible transitions have a change in energy of 4 eV. Emitted Photon Energies (left to right) 1 eV, 3 eV, 6 eV, 2 eV, 5 eV, 3 eV (E) (B) (A) Which of the following transitions will produce the photon with the longest wavelength? (A) n = 2 to n = 1 (B) n = 3 to n = 1 (C) n = 3 to n = 2 (D) n = 4 to n = 1 (E) n = 4 to n = 3 (C) (D) Longest wavelength means least energy. (Large wavelength, low frequency, less energy.) The lowest energy photon will be emitted during the transition that has the smallest ΔE for the electron. (E) (B) (A) Which of the following transitions will produce the photon with the longest wavelength? (A) n = 2 to n = 1 (B) n = 3 to n = 1 (C) n = 3 to n = 2 (D) n = 4 to n = 1 (E) n = 4 to n = 3 (C) (D) Longest wavelength means least energy. (Large wavelength, low frequency, less energy.) The lowest energy photon will be emitted during the transition that has the smallest ΔE for the electron. Atoms can emit photons if they drop to a lower energy level, or absorb energy and raise to a higher energy level. However, they must do this is the exact perfect amount that will bring them to an allowed energy state! Mystery: Solved. Final Whiteboard: Absorption and Emission The energy level diagram below is for a hypothetical atom. A gas of these atoms initially in the ground state is irradiated with photons having a continuous range of energies between 7 and 10 electron volts. One would expect photons of which of the following energies to be emitted from the gas? (A) 1, 2, and 3 eV only (B) 4, 5, and 9 eV only (C) 1, 3, 5, and 10 eV only (D) 1, 5, 7, and 10 eV only (E) Since the original photons have a range of energies, one would expect a range of emitted photons with no particular energies. How to think about this The energy level diagram below is for a hypothetical atom. A gas of these atoms initially in the ground state is irradiated with photons having a continuous range of energies between 7 and 10 electron volts. If the photons transferred their energy to the atom, … could jump to any state in this range! Electrons starting in this state… Since the only allowed energy level within that range is the second excited state (-5 eV), the electrons in the atom will be excited to that state! After the electrons are excited, they fall back down to lower energy levels During this time, photons of energy -ΔEelectron are emitted! 9 eV 5 eV 4 eV (A) 1, 2, and 3 eV only (B) 4, 5, and 9 eV only (C) 1, 3, 5, and 10 eV only (D) 1, 5, 7, and 10 eV only (E) Since the original photons have a range of energies, one would expect a range of emitted photons with no particular energies. Something to Think About for Next Time But why does an electron only have distinct allowed atomic energies? What prevents the atom from collapsing still? Wouldn’t the electrons and protons attract and there be a certain probability of them annihilating one another? Atomic Theory Wrap-up and Review Announcements Test Monday – Nuclear Physics, The Photoelectric Effect, Photons and Matter, Atomic Physics 20 MC – All topics 2 FRQs – 1 on nuclear physics, 1 on atomic energy levels and de Broglie wavelength AP Test Takers! You have mandatory AP TEST GRIDDING during 8th period Monday in the Cafeteria. Question 1c) from Homework The atom is in the ground state when an electron traveling with a speed of 1.3 x 106 m/s collides with it. Can the electron excite the atom to the n = 2 state? ____Yes ____No ____It cannot be determined. Justify your answer. n=3 n=2 n=1 Does the incoming electron have enough KE to give the atom the 3 eV it needs to jump to the next energy state? (Hint: The incoming electron does not need to transfer all of its KE!) Question 1c) from Homework The atom is in the ground state when an electron traveling with a speed of 1.3 x 106 m/s collides with it. Can the electron excite the atom to the n = 2 state? ____Yes ____No n=3 ____It cannot be determined. The incident electron has ½ mv2 = 4.8 eV of KE. The atom needs 3 eV to jump. n=2 An incident electron, unlike an incident photon, can transfer only some of its energy to the atom. n=1 Follow-Up Question The atom is in the ground state when photons with energies that range in energies from 4.5-5.5 eV are incident upon it. To what excited state(s), if any, can the atom be excited? n=3 n=2 n=1 An electron must absorb the full energy of an incident photon. No fractions or combinations! This means that the atom will be able to be excited to the n = 3 state. E photon = DEelectron Energy Level Diagram – Part I An experiment is performed on a sample of atoms known to have a ground state of -5.0 eV. The gas is illuminated with “white light” (400 – 700 nm). A spectrometer capable of analyzing radiation in this range is used to measure the radiation. The sample is observed to absorb light at only 400 nm. After the “white light” is turned off, the sample is observed to emit visible radiation of 400 nm and 600 nm. (a)In the space below, determine the values of the energy levels and on the following scale sketch an energy-level diagram showing the energy values in eV and the relative positions of: i. The ground state ii. The energy level to which the system was first excited. iii. One other energy level that the experiment suggests may exist An experiment is performed on a sample of atoms known to have a ground state of -5.0 eV. The gas is illuminated with “white light” (400 – 700 nm). A spectrometer capable of analyzing radiation in this range is used to measure the radiation. The sample is observed to absorb light at only 400 nm. E photon = hc l 400 nm light possesses 3.1 eV of energy. State to which atom was first excited +3.1 eV It can raise the atom to a -1.9 eV level. n = 1 (ground state) After the “white light” is turned off, the sample is observed to emit visible radiation of 400 nm and 600 nm. 400 nm is expected! If the atom absorbed 400 nm light to jump from -5 eV to -1.9 eV, then it should also emit 400 nm light when it jumps back from -1.9 eV to -5 eV 600 nm light possesses 2.1 eV of energy. In order for the atom to emit 600 nm light once it has been excited, there must be an energy level 2.1 eV below the excited state to which the atom was first raised. n=3 -2.1 eV -3.1 eV n=2 n=1 Atomic Physics - Part II! What is the wavelength of any other radiation, if any, that might have been emitted in the experiment? Why was it not observed? n=3 n=2 n=1 Solution - Part II! What is the wavelength of any other radiation, if any, that might have been emitted in the experiment? What was it not observed? Given the energy levels that we drew, there should be a photon emitted during the n = 2 to n = 1 transition. n=3 n=2 -1 eV n=1 The energy emitted during this transition is 1 eV λ = 1,238 nm Infrared But we still have one major unanswered question: Why are electrons in atoms only “allowed” to occupy certain specific orbital energies? Is this something totally unique, or can we come up with a deeper reason for why this happens? By combining his theory with de Broglie’s hypothesis, Niels Bohr figured out the reason why only certain energy level orbits are allowed! Review: deBroglie Wavelength On a quantum scale, all matter must be modeled by using wave functions, which allow you to predict the probability of finding the particle in a particular region of space. The deBroglie wavelength, λ = h/p, defines the “size” of the particle’s wave function. Electron traveling at 103 m/s Baseball traveling at 103 m/s Orbital Standing Waves Bohr theorized that the “allowable” orbital electron energies were the ones in which the circumference of the orbit is a multiple of the deBroglie wavelength! Bohr figured out that the electron’s orbit would only be stable if its circumference corresponded to a standing wave with the electron’s deBroglie wavelength! Think of wrapping a sine wave around a circle, and only allow the ones that have an exact whole number of waves around the circumference. An electron’s orbit will only be stable if the circumference corresponds to a standing wave with the electron’s deBroglie wavelength! Think of wrapping a wave around the edge of a circle. This would the n = 3 state (second excited state)! Ground state is the lowest possible energy This is why an electron cannot get infinitely close to the nucleus. The closest possible orbital that the electron can have corresponds to one full deBroglie wavelength around the outside of the orbit. An experimenter determines that when a beam of monoenergetic electrons bombards a sample of pure gas, atoms of the gas are excited if the kinetic energy of each electron in the beam is 3.70 eV or greater. a)Determine the deBroglie wavelength of 3.70 eV electrons. b)Once the gas is excited by 3.70 eV electrons, it emits monochromatic light. Determine the wavelength of this light. Experiments reveal that two additional wavelengths are emitted if the beam energy is raised to at least 4.90 eV. c) In the space below construct an energy-level diagram consistent with this information and determine the energies of the photons associated with those two additional wavelengths. a)First, use K = ½ mv2 to solve for velocity: 1,140,000 m/s – then p = mv λ = h/p gives 0.64 nm. b) E = hc/λ gives λ = 334 nm (ultraviolet light) c) n=3 1.2 eV n=2 4.9 eV 3.7 eV n=1 Using E = hc/λ, we obtain that the other two wavelengths emitted are 103 nm and 253 nm. Photoelectric Effect Review Light of wavelength 400 nm is incident on a metal surface. Electrons are ejected from the surface with maximum kinetic energy 1.1 x 10-19 J. a) Calculate the frequency of the incoming light. b) Calculate the work function of the metal surface. c) Calculate the stopping potential for the emitted electrons. d) Calculate the momentum of an electron with the max kinetic energy. Kmax = hf - F Solution Kmax = eDVstop a) Just use c = λf! Part (a) of these FRQs is usually a simple plug-and-chug. Don’t overthink it! Answer = b) Use Kmax = hf – Φ to get Φ = c) Use Kmax = qeΔVstop! d) K = ½ mev2 allows you to solve for velocity. You can then plug into p = mv to solve for momentum. Equations to Memorize Nuclear Physics DE = Dmc 2 Photon Model c=lf E = hf = hc l Photoelectric Effect p=h l Kmax = hf - F deBroglie Wavelength Kmax = eDVstop fthreshold F = h l=h p Atomic Physics E photon = DEelectron