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- 1. Today’s Lecture: Bohr AtomicModel You must be able to understand atomic structure. 1
- 2. 2 The electron inthe hydrogen atom can occupy only certain specific circular orbits of quantized angular momentum and no others. the angular momentum in an allowed orbit must be an integer n (the quantum number) times h/2π, where h is Planck's constant. Since the electron of mass m moves with a speed v in a circular path of radius r, its angular momentum is L = mvr. this condition is Ln = rnmvn = nh/2π Bohr Atomic Model Niels Bohr Nobel Prize 1922 “the structure of atoms and the radiation emanating from them”
- 3. • de Brogliecombined the condition for standing waves, nλ = 2πr, with his expression for the wavelength of matter, λ = h/p. Since the linear momentum is p = mv, the standing wave condition can be written as follows: nλ = 2πr nh/mv = 2πr • A simple rearrangement yields rmv = nh/2π. This is precisely Bohr's orbital condition. • Thus, matter wave are seen to have a direct connection to physics on the atomic level. Quantized orbits or angular momentum
- 4. Light is emittedwhen an electron jumps from a higher orbit to a lower orbit and absorbed when it jumps from a lower to higher orbit. The energy and frequency of light emitted or absorbed is given by the difference between the two orbit energies, e.g., E(photon) = E2 - E1 (Energy difference) hʋ= E2 - E1 (Energy difference) Bohr used these observations to argue that the energy of a bound electron is "quantized." E1 E2 E2 E1 Photon absorbed Photon emitted + h - h Photon absorbed Photon emitted Bohr Atomic Model
- 5. Light absorption occurswhen an electron absorbs a photon and makes a transition for a lower energy orbit to a higher energy orbit. Absorption spectra appear as sharp lines. Eni - Enf = h Photon Absorbed E5 E4 E3 E2 E1 The Bohr Atom 1 2 3 4 5 E5 - E1 E4 - E1 E3 - E1 E2 - E1
- 6. Light emission occurswhen an electron makes a transition from a higher energy orbit to a lower energy orbit and a photon is emitted. Emission spectra appear as sharp lines. Eni - Enf = h Photon Emitted E3 E2 E1 E5 E4 The Bohr Atom 1 2 3 4 5 E5 - E1 E4 - E1 E3 - E1 E2 - E1
- 7. Sample Problem • Calculatethe energy of the photon that is emitted when a hydrogen atom changes from energy state n=3 to n=2. What color corresponds with the photon emitted? • Solution • From reference tables… E3 = Einitial = -1.51 eV E2 = Efinal = -3.40 eV Ephoton = Einitial – Efinal = (-3.40 eV) – (-1.51 eV) = -1.89 eV Ephoton = hf ==== f = Ephoton / h But we need Ephoton in Joules, because Planck’s constant is in Joules Ephoton = (-1.89 eV) (1.60x10-19J / eV) = 3.02x10-19 J f = (3.02x10-19 J) / (6.63x10-34 J∙s) = 4.56x1014 Hz From reference tables, this frequency corresponds with red light.
- 8. 8 Summary of theBohr Model – The electron in the hydrogen atom can occupy only certain specific orbits of fixed radius and no others. – The electron can jump from one orbit to a higher one by absorbing a quantum of energy in the form of a photon. – Each allowed orbit in the atom corresponds to a specific amount of energy. The orbit nearest the nucleus represents the smallest amount of energy that the electron can have.
- 9. 9 1. It cannotexplain the spectra of atoms or ions having more than one electron. 2. High resolution spectroscopy has shown that some spectral lines in fact consist of a group of closely spaced lines. Bohr’s theory cannot explain this fine structure of spectral lines. 3. Theoretically, the choice of circular, rather than elliptic, orbits is an over-simplification. 4. It is now known that the planetary model of the atom is not a correct representation because the orbits of electrons cannot be precisely defined. Limitations of Bohr Theory
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