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CHAPTER 6 ELECTRONIC STRUCTURE OF ATOMS

CHAPTER 6 ELECTRONIC STRUCTURE OF ATOMS. Almost everything we now know about electrons came from the study of light. Light is a type of electromagnetic radiation. Types of Electromagnetic Radiation. Three complete cycles of wavelength (l ).

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CHAPTER 6 ELECTRONIC STRUCTURE OF ATOMS

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  1. CHAPTER 6 ELECTRONIC STRUCTURE OF ATOMS

  2. Almost everything we now know about electrons came from the study of light. Light is a type of electromagnetic radiation.

  3. Types of Electromagnetic Radiation

  4. Three complete cycles of wavelength (l)

  5. The product of the frequency (n), and its wavelength (l) equals the speed of light (c). c = ln

  6. C =  Since the speed of light is 3 x 108 m/s, there must be an inverse relationship between wavelength and frequency

  7. The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is the frequency of this radiation?

  8. The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is the frequency of this radiation? c =lnn = 5.09 x 1014 s-1

  9. Max Planck assumed that small quantities of energy (quantum) can be emitted or absorbed as electromagnetic radiation. Planck’s constant (h) = 6.63 x 10-34 joule-seconds (J-s) E = hv

  10. Since yellow light has a wavelength of 589 nm, what is the smallest quantum of energy that can be absorbed from yellow light?

  11. Since yellow light has a wavelength of 589 nm, what is the smallest quantum of energy that can be absorbed from yellow light? E=hv E=(6.63 x 10-34)(5.09 x 1014) E = 3.37 x 10-19 J

  12. Johann Balmer observed the 4 lines of the hydrogen spectrum to come up with a formula: v=C(1/22-1/n2) n= 3,4,5,6

  13. Niels Bohr assumed that electrons orbit in circular paths around the nucleus. He also said that orbits of certain radii correspond to certain energies.

  14. En=(-RH)(1/n2) n= 1, 2, 3, 4, … n= the principal quantum number RH = 2.18 x 10-18 J

  15. Bohr assumed that electrons could “jump” from one energy state to another by absorbing or emitting energy of certain frequencies. E= Ef - Ei = hv

  16. By combining some equations: v= E/h = (RH/h)(1/ni2-1/nf2) Calculate the wavelength of light that corresponds to the transition of the electron from the n=4 to n=2 state of the H atom.

  17. Heisenberg’s UncertaintyPrinciple: It is impossible for us to know the exact velocity or exact location of an electron

  18. This is a probability density of the location of the electron in a H atom. This represents the shape of the s orbital.

  19. The p Orbitals

  20. The d Orbitals

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