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Calculating particle properties of a wave

Calculating particle properties of a wave. A light wave consists of particles (photons): The energy E of the particle is calculated from the frequency f of the wave via Planck : E = h f ( 1) A particle can act like a wave:

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Calculating particle properties of a wave

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  1. Calculating particle properties of a wave • A light wave consists of particles (photons): The energy E of the particle is calculated from the frequency f of the wave via Planck: E = hf (1) • A particle can act like a wave: The momentum pof the particle is calculated from the wavelength  via de Broglie: p = h/ (2) • Frequency f and wavelength  can be conver- ted into each other by the wave equation:f = v(3) Ch. 12

  2. · · · · · · = 2.5 eV A practical energy unit in quantum physics is an eV, the energy of an electron accelerated by 1 Volt: eV = electron-Volt = (electron charge e ) · (1 Volt) = (1.6·10-19Coulomb) · (1 Volt) = 1.6·10-19 Joule Conversion of Jouleto eV: J = 6.24·1018 eV Energy of a photon • Consider light with a wavelength =500 nm • Use (1),(3), v=c (1) (3)

  3. Conversion between wavelength and energy Wavelength and energy areinversely proportional with the proportionality constant hc: E = hc/ = hc/E hc= 1240 eV ·nm Green light for example: E = 1240eV·nm / 550nm = 2.25eV

  4. 10-3 s ·4·10-14 W = 4·10-17 J = 6.24·1018 ·4·10-17 eV = 250eV; 250eV/2.25eV = 111 photons How many photons can you see? In a test of eye sensitivity, experimenters used 1 millisecond flashes of green light. The lowest light power that could be seen was 4·10-14Watt. How many green photons (550 nm, 2.25 eV) is this? • 10 photons • 100 photons • 1,000 photons • 10,000 photons

  5. Detecting single photons • A photon can be converted into a pulse of electrons: • For ultraviolet light, X-rays, Gamma rays: In aGeiger counter a photon knocks electrons from gas molecules, creating a miniature spark. • For visible light: In aphotomultiplier a photon kicks an electron out of a solid (the photocathode). The electron is multiplied into amillionelectronsbymultiplebounces.

  6. Kamiokande particle detector lined with photomultipliers Phy107 Lecture 16

  7. 6.6 s · · ev · The wave character of electrons • The wavelength is inversely proportional to the velocity v of the electron and to its mass me. • Anelectronwithlowvelocity(lowkineticenergy) has a long wavelength,which is easier to detect. Ch. 12.4

  8. Wavelength of an electron from its kinetic energy 1) Velocity vfrom the kinetic energy Ekin: Ekin= ½mev2v=2Ekin/me • Wavelength  via deBroglie and p=mev: = h/mev For a kinetic energy Ekin = 1eVone obtains: 1)v= 2·1.6·10-19 J / 9·10-31 kg (eV =1.6·10-19J) =6·105m/s(me =9·10-31 kg) 2) = 6.6·10-34Js /9·10-31 kg6·105m/s(h =6.6·10-34Js) =1·10-9 m = 1 nm

  9. Typical wavelength of an electron • The wavelength of an electron is very short and hard to measure with macroscopic tools. • The spacing of nickel atoms in a crystal was needed to demonstrate electron diffraction. • A typical electron energy in a solid with 1eVenergyhasawavelengthofonly1nanometer. • Nanotechnologyhas the capability to shape electron waves and thereby influence the be-havior of electrons in a solid (Lect. 24a).

  10. v · · · · · Wavelength of a football NFL guidelines: “Weight 14to15 ounces” = 0.4 kg Brett Favre can throw the ball 70 miles/hour = 30 m/s That is similar to the Planck length where space-time falls apart, 1014 times smaller than the smallest measurable length (Lect. 3). Conclusion: Macroscopic objects don’t show measurable quantum effects.

  11. When are quantum phenomena important?(optional) 1) Quantum physics is important atsmall distances d,smaller than the deBroglie wavelength =h/p: d <  Example: Nanotechnology for shaping electron waves. 2) Quantum physics is important for large energy quanta E = hf: Example: Planck’s radiation law cuts the spectrum off when the energy to create a photon exceeds the available thermal energy (Etherm 0.1 eV at T=300K): E > Etherm Example: The photoelectric effect occurs only when the photon energy exceeds the minimum energy  to knock out an electron (  = “work function”4eV ): E > 

  12. Criteria 1) and 2) are connected by the wave equation:  · f = v ( v= velocity of the wave ) If the wavelength  is small the frequency fis largeand the energy quantum E=hfis large. Example: Quantum computingrequires that quantum effects are dominant. Three approaches to quantum computing are pursued in the Physics Department (silicon quantum dots, superconducting junctions, and cold atoms in traps).All of them use very low temperatures to keep the quantum effects undisturbed by thermal noise (Criterion 2).

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