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Electrons as Waves & Particles

Ch. 5 - Electrons in Atoms. Electrons as Waves & Particles. Quantized Energy comes in discrete packages Example : second hand on clock that “ticks” STAIRS. Continuous Energy is flowing Example : second hand on clock that moves continuously ESCALATOR.

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Electrons as Waves & Particles

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  1. Ch. 5 - Electrons in Atoms Electrons as Waves & Particles

  2. Quantized Energy comes in discrete packages Example: second hand on clock that “ticks” STAIRS Continuous Energy is flowing Example: second hand on clock that moves continuously ESCALATOR Quantized Energy vs. Continuous Energy

  3. Dual Nature of Light…….Particle or Wave • Remember a quantum of energy is the amount of energy to move an electron from one energy level to another. • Energy is quantized therefore light must be quantized. • These smallest pieces, quanta, are called ……photons: particles of light • BUT, Energy is also continuous. Therefore light which is continuous acts like a WAVE

  4. Therefore………. Light transmits energy as a particle And Light travels through space as a wave

  5. Quantum Theory Einstein (1905) Concluded - light has properties of both waves and particles “ wave - particle duality ”

  6. Wave-Particle Duality JJ Thomson won the Nobel prize for describing the electron as a particle. His son, George Thomson won the Nobel prize for describing the wave-like nature of the electron. The electron is a particle! The electron is an energy wave!

  7. LIGHT • Wave-Particle Duality • JJ Thomson won the Nobel prize for describing the electron as a particle. • His son, George Thomson won the Nobel prize for describing the wave-like nature of the electron. • The electron is a particle! • The electron is an energy wave!

  8. Electromagnetic Radiation“Light” • The study of light led to the development of the quantum mechanical model. • Electromagnetic radiation includes many kinds of waves • Light is a type of electromagnetic radiation. • All EM waves move at 3.00 x 108 m/s (c =the Speed of Light)

  9. EM Spectrum HIGH ENERGY LOW ENERGY

  10. The Electromagnetic Spectrum

  11. ElectromagneticSpectrum In Increasing Energy: ROY G BIV

  12. Electromagnetic Spectrum

  13. LIGHT

  14. A. Waves • Wavelength () - length of one complete wave • Frequency () - # of waves that pass a point during a certain time period • hertz (Hz) = 1/s or s-1 • Amplitude (A) - distance from the origin to the trough or crest

  15. Crest Wavelength Amplitude Trough Parts of a wave High point Origin baseline of wave Low point

  16. Wavelength – distance from crest to crest • symbol: λ= “lambda” Amplitude – height of wave from the origin to the peak; brightness, intensity of light

  17. Frequency – how frequentlya way oscillates up & down; the # of times a wave completes a cycle of up & down motion Symbol is ν=“nu” SI unit isHertz (Hz)orcycles/sec (1s or s-1)

  18. Relationship between Frequency & Wavelength DECREASES As Wavelength INCREASES, frequency ________________ INCREASES As Wavelength DECREASES, frequency _______________

  19. Speed of Light – 3 x 108 m/s Frequency c = ln Wavelength

  20. EM Spectrum • Frequency & wavelength are inversely proportional c =  c: speed of light (3.00  108 m/s) : wavelength (m, nm, etc.) : frequency (Hz, 1/s or s-1)

  21. Energy of a wave – E (measured in joules) Frequency E = hn Planck’s Constant 6.626 x 10-34 j*s

  22. EM Spectrum • The energy of a photon is proportional to its frequency. E: energy (J, joules) h: Planck’s constant (6.6262  10-34 J·s) : frequency (s-1) E = h

  23. Summary of Light Therefore: wavelength and frequency are indirectly proportional. Short wavelength = high frequency Long wavelength = low frequency c = ln Therefore: energy is directly proportional to the frequency. E = hn High frequency = high energy Low frequency = low energy

  24. WORK:  = c   = 3.00  108 m/s 4.34  10-7 m EM Spectrum • EX: Find the frequency of a photon with a wavelength of 434 nm. GIVEN:  = ?  = 434 nm = 4.34  10-7 m c = 3.00  108 m/s = 6.91  1014 s-1

  25. EM Spectrum • EX: Find the energy of a red photon with a frequency of 4.57  1014 s-1. GIVEN: E = ?  = 4.57  1014 s-1 h =6.6262  10-34 J·s WORK: E = h E = (6.6262  10-34 J·s) (4.57  1014 s-1) E = 3.03  10-19 J

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