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Vacuum tube - V, only for shorter than certain wavelength Current V VoVo Fixed wavelength Varying intensity I2I 3I Maximum electron energy 0

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Current V 1 2 3 1 < 2 < 3 Q.Why the maximum electron energy depend on the wavelength (frequency)? Why strong beams of long wavelength cannot knock out electrons? Classical EM theory (wave theory) tells us that the max electron energy should depend on only the intensity of the light. In 1900, Max Planck postulated that electromagnetic energy is emitted in discrete packets, or quanta. The energy of these quanta is proportional to the frequency of the radiation. E = hf = hc/ Huge water wave, large amplitude sound wave h = 6.6 x 10 -34 Js (Planck’s constant)

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In explaining the photoelectric effect, Einstein picked up the idea of Planck, and proposed in 1905 that light was not only emitted by bundles of energy E = hf, but it was also absorbed in such bundles. Light Quanta = Photons Kinetic Energy of an electron = E photon - E threshold E photon = hc/ > E threshold Work function: depends on material ~ eV - light quantum: photon

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In this case, light can be considered as a massless particle, photon with energy solely determined by its wavelength (frequency). There is no conflict with Einstein’s relativistic mechanics: massless particle can have speed of light! E = hf = hc/ h = 6.6 x 10 -34 Js (Planck’s constant) = 580 nm photon (yellow light) carries E = (6.6 x 10 -34 Js)(3 x 10 8 m/s)/(580 x 10 -9 m) = 3.4 x 10 -19 J

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How do we understand the intensity of light? Intensity of light (EM radiation) ≈ number of photons with same wavelength or frequency Wave behaves like a particle. Does a particle behaves like wave?

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Q. Light of wavelength 650 nm is required to cause electrons to be ejected barely from the surface of a particular metal. What is the kinetic energy of the ejected electrons if the surface is bombarded with light of wavelength 450 nm? “…ejected barely…” means that ejected electron has ≈ 0 kinetic energy. Kinetic Energy of an electron = E photon - E threshold h /c = E threshold (work function) for = 650 nm E threshold = (6.6 x 10 -34 Js)(3 x 10 8 m/s)/(659 x 10 -9 m) = 3.06 x 10 -18 J E k = (6.6 x 10 -34 Js)(3 x 10 8 m/s)/(450 x 10 -9 m) – 3.06 x 10 -18 = (4.42 – 3.06) x 10 -18 = 1.36 x 10 -19 J

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A light source produces 400 nm wavelength. When the light strikes a metal surface, a stream of electrons emerges from the metal. If the intensity of the light is doubled, 1. the electrons emitted are more energetic. 2. more electrons are emitted in a given time interval 3. both of the above 4. none of the above

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In this case, light can be considered as a massless particle, photon with energy solely determined by its wavelength (frequency). There is no conflict with Einstein’s relativistic mechanics: massless particle can have speed of light! E = hf = hc/ h = 6.6 x 10 -34 Js (Planck’s constant) = 580 nm photon (yellow light) carries E = (6.6 x 10 -34 Js)(3 x 10 8 m/s)/(580 x 10 -9 m) = 3.4 x 10 -19 J Wave behaves like a particle. Does a particle behaves like wave?

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It would seem that the basic idea of the quantum theory is the impossibility of imagining an isolated quantity of energy without associating with it a certain frequency de Broglie in 1923 as a graduate student light quantum (photon) E = hf = hc/ Relativistic case massless E = pc = hc/ = h/p wave quantity particle quantity for m o = 0

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= h/p De Broglie proposed that all particles (electrons) should have wavelength associated with their momentum in exactly the same manner. particlewave

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Q. What is the de Broglie wavelength of an electron that has a kinetic energy of 100 eV? After an electron is accelerated in 100 V potential difference, its kinetic energy is 100 eV. eV unit has to be converted into SI unit, Joule. 1 eV = 1.6 x 10 -19 J E k = (1/2)m o v 2 = 1.6 x 10 -17 J v 2 = 2E k /m o = 2(1.6 x 10 -17 J)/(9.1 x 10 -31 kg) = 3.52 x 10 13 m 2 /s 2 v = 5.93 x 10 6 m/s low speed: no need to use relativistic = h/p = h/m o v = (6.6 x 10 -34 Js)/(9.1 x 10 -31 kg x 5.93 x 10 6 m/s) = 1.23 x 10 -10 m = 0.123 nm

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C.J. Davisson and L.H Germer scattered low energy electrons off The Ni crystal and observed a peculiar pattern of scattered electrons. Diffraction pattern from the regular Crystalline structure of order of 0.1 nm size.

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Hydrogen Spectrum Emission SPECTRUM ABSORPTION SPECTRUM

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Bohr’s Hydrogen Atom and the Electron as a Wave Niels Bohr (1885 – 1962) Louis de Broglie (1892 – 1987) In 1913, Rutherford’s atom received a quantitative description from Niels Bohr who explained experimentally observed discrete nature of atomic spectrum of Hydrogen. In spite of its immediate success in providing theoretical account of the spectrum and other nature of Hydrogen atom, a complete understanding of Bohr’s atom came only after de Broglie’s conjecture (1923) that electrons should display wave properties. + -

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Bohr’s Hydrogen Atom (1 proton and 1 electron) + - - - centripetal force = Coulomb force electron as a wave (de Broglie) 2 r =, 2, 3, … = h/mv This is the necessary condition for electron to maintain an orbit.

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2 r =, 2, 3, … = h/mv 2 r n = nh/mv nn r n = (5.3 x 10 -11 )n 2 (m); v n = 2 x10 6 /n (m/s)n r n (nm) v n (m/s) 10.053 2 x 10 6 20.212 1 x 10 6 30.477 5 x 10 5 2 x 0.053 0.1 nm is the size of Hydrogen atom

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Now, let’s think about the total energy of the electron in the n th orbit. E tot = Potential Energy + Kinetic Energy -ke 2 /r n (1/2)mv n 2n E n (eV) 1-13.6 2-3.4 3-1.5 ∞0

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+ - - - E tot 0 n = ∞ -13.6 eVn = 1 n = 2-3.4 eV n = 3-1.5 eV - Ionized state of Hydrogen: proton

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Electron Energy diagram of Hydrogen atom E tot 0 n = ∞ n = 2-3.4 eV n = 3-1.5 eV n = 1-13.6 eV - lowest energy: ground state Electron has to absorb 10.2 eV energy for this - Electron has to absorb 12.1 eV energy for this

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in eV (Rydberg const.) E(n m) or < 0Absorb photons of a given E (n m) or > 0Emit photons of a given E = hc/

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E tot 0 n = ∞ n = 2-3.4 eV n = 3-1.5 eV n = 1-13.6 eV - E (1 2) = 10.2 eV = 10.2 x (1.6 x 10 -19 J/eV) = 1.63 x 10 -18 J E = hc/ = 121 nm Ultraviolet range

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Q What is the longest wavelength em radiation that can ionize unexcited hydrogen atom? smallest energy ground state (n = 1) n = 1 m = ∞ 1/ = (1.097 x 10 7 ) x (0 – 1) = - 1.097 x 10 7 (m -1 ) = -9.12 x 10 -8 (m) = -91.2 (nm) (- means absorption)

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Q. What is the shortest wavelength for the Balmer series? Largest energy difference in Balmer series From n = ∞ to m = 2 (Balmer) = (R x (1/4 – 0)) -1 = 365 nm The Balmer series are in the visible range.

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