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Light

Explore the relationship between wavelength, frequency, and energy in light waves, and learn how to calculate these properties using Plank's constant. Understand the atomic emission spectrum and the energy levels of electrons in different states. Discover how the energy released by electrons corresponds to the color of light observed.

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Light

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  1. Light Waves and Particle Characteristics

  2. Parts of a Wave l = wavelength (lambda) n=frequency(nu)

  3. Wave Wavelength (m) – Longer wavelengths have lower energy than shorter wavelengths. We say that l is inversely proportional to E l a 1/E Frequency (1/sec) – As the wavelength decreases, the frequency increases We say that n is inversely proportional to l n a 1/l

  4. Waves We can use the relationship c=ln to solve for either the wavelength or frequency, when the other is known. C = speed of light – This is constant in a vacuum 3.00 x 108m/sec

  5. Example #1 Calculate the wavelength of yellow light emitted by a sodium vapor lamp if the frequency of the radiation is 5.10 x 1014Hz? (Hz = 1/sec) C = ln 3.00 x 108m/s = l(5.10 x 1014Hz) 5.88 x 10-7m = l

  6. Example #2 What is the wavelength of radiation with a frequency of 1.50 x 1013Hz? Does it have a longer or shorter wavelength than red light? c = ln 3.00 x 108m/s = l(1.50 x 1013Hz) 2.00 x 10-5m = l It has a longer wavelength than red light

  7. Light Each element has a specific Atomic Emission Spectrum – as specific frequencies of light are emitted What causes the frequencies of light?

  8. Light As e- gain Energy, they move to higher energy levels, away from the nucleus They are moving from a Ground State to an Excited State As the e- move back down to Ground State – that Energy must be released

  9. Light The amount of Energy that is absorbed or released is known as a Quantum of Energy Energy is proportional to n E a n E = hn where h = Plank’s Constant

  10. Light Plank’s Constant = 6.6262 x 10-34 Jsec J = Joule (a unit of energy) E ( ) = h (J sec) n (1/sec) E (J)

  11. Light The greater the jump, the greater the amount of energy that will be released n=6 n=1 released more energy than an e- which moves from n=6 n=5

  12. Example #3 How much energy is released by an e- dropping from n=4 n=2 if the light that is emitted has a frequency of 5.55 x 1014 1/sec? E = hn E= 6.6262 x 10-34Jsec (5.55 x 1014 1/sec) E = 3.68 x 10-19 J

  13. Example #4 If an electron releases 4.56 x 10-19J of energy as it drops from n=5 n=2, what color of light would be observed?

  14. Example 4 continued Begin with E=hn 4.56 x 10-19J = 6.6262 x 10-34Jsec (n) 6.88 x 1014 1/sec = n Next use c = ln 3.00 x 108m/sec = l (6.88 x 1014 1/sec) 4.37 x 10-7m = l Blue light is observed

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