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Arrangement of the Electrons Chapter 4 (reg.)

Arrangement of the Electrons Chapter 4 (reg.). (Electron Configurations). Spectrum of Light!. Electromagnetic Radiation -form of energy that exhibits wave-like behavior as it travels through space.

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Arrangement of the Electrons Chapter 4 (reg.)

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  1. Arrangement of the Electrons Chapter 4 (reg.) (Electron Configurations)

  2. Spectrum of Light! • Electromagnetic Radiation-form of energy that exhibits wave-like behavior as it travels through space. • Electromagnetic Spectrum-ordered arrangement by wavelength or frequency for all forms of electromagnetic radiation.

  3. Parts of the wave • Wavelength-lambda (λ) The distance between corresponding points on adjacent waves. Units: m, nm, cm, or Å • Frequency-nu (ν) The number of waves passing a given point in a definite amount of time. Units: hertz (Hz) or cycles/sec = 1/sec = sec-1

  4. When an electric field changes, so does the magnetic field. The changing magnetic field causes the electric field to change. When one field vibrates—so does the other. • RESULT-An electromagnetic wave.

  5. Waves or Particles • Electromagnetic radiation has properties of waves but also can be thought of as a stream of particles. • Example: Light • Light as a wave: Light behaves as a transverse wave which we can filter using polarized lenses. • Light as particles (photons) • When directed at a substance light can knock electrons off of a substance (Photoelectric effect)

  6. Relationship between λ and ν • c = λ∙ν • λ = wavelength (m) • ν = frequency (Hz) • c = speed of light= 3.0 x 108 m/sec (constant) • λ and ν are _______________ related.

  7. Practice Problem Truck-mounted helium-neon laser produces red light whose wavelength (λ ) is 633 nanometers. Determine the frequency (v). *Remember that c=3.0x108m/s. *Use the formulav= c λ

  8. c= λ .v c =3.0x108 m/s c= λ .v v=c / λ λ = 633nm= 6.33x10-7m v = 3.0x108 m/s = 0.47x 1015s-1 = 4.7x1014 s-1 6.33x10-7m Frequency = 4.7x1014 Hz (cyclesper second)

  9. WORK:  = c   = 3.00  108 m/s 4.34  10-7 m B. 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 Hz

  10. Light as waves and particles(the Particle Theory of light) • 2 problems that could not be explained if light only acted as a wave. • 1.) Emission of Light by Hot bodies: Characteristic color given off as bodies are heated: red  yellowwhite If light were a wave, energy would be given off continually in the infrared (IR) region of the spectrum.

  11. The second problem……… • 2.) Absorption of Light by Matter = Photoelectric Effect Light can only cause electrons to be ejected from a metallic surface if that light is at least a minimum threshold frequency . The intensity is not important. If light were only a wave intensity would be the determining factor, not the frequency!

  12. Max Planck (1900’s)Particle Theory of Light • When an object loses energy, it doesn’t happen continuously but in small packages called “quanta”. “Quantum”-a definite amount of energy either lost or gained by an atom. “Photon”-a quantum of light or a particle of radiation.

  13. Calculate the frequency for the yellow-orange light of sodium. • Calculate the frequency for violet light.

  14. Calculate the frequency for the yellow-orange light of sodium. • Calculate the frequency for violet light.

  15. Relationship betweenEnergy and ν • E = h∙ν • E = energy (joule) • h = Planck’s constant = 6.63 x 10-34j∙sec • ν = frequency (Hz) • E and ν are ______________ related. • Calculate the energy for the yellow-orange light for sodium. • Calculate the energy for the violet light.

  16. Line Spectrums • Excited State: Higher energy state than the atom normally exists in. • Ground State: Lowest energy state “happy state” • Line Spectrum: Discrete wavelengths of light emitted. • 2 Types: • 1.) Emission Spectrum: All wavelengths of light emitted by an atom. • 2.) Absorption Spectrum: All wavelengths of light that are not absorbed by an atom. This is a continuous spectrum with wavelengths removed that are absorbed by the atom. These are shown as black linesfor absorbed light. • Continuous Spectrum: All wavelengths of a region of the spectrum are represented (i.e. visible light)

  17. Hydrogen line Spectrum & niel’s Bohr • Hydrogen’s spectrum can be explained with the wave-particle theory of light. • Niel’s Bohr (1913) • 1.) The electron travels in orbits (energy levels) around the nucleus. • 2.) The orbits closest to the nucleus are lowest in energy, those further out are higher in energy. • 3.) When energy is absorbed by the atom, the electron moves into a higher energy orbit. This energy is released when the electron falls back to a lower energy orbit. A photon of light is emitted.

  18. Hydrogen Spectrum • Lyman Series-electrons falling to the 1st orbit, these are highest energy, _____ region. • Balmer Series- electrons falling to the 2nd orbit, intermediate energy, _______ region. • Paschen Series-electrons falling to the 3rd orbit, smallest energy, ______ region.

  19. Bohr’s equation for Hydrogen • En = (-RH) 1/n2 • En = energy of an electron in an allowed orbit (n=1, n=2, n=3, etc.) • n = principal quantum number (1-7) • RH = Rydberg constant (2.18 x 10-18 J) • When an electron jumps between energy levels: ΔE =Ef – Ei • By substitution: ΔE = hν = RH(1/ni2 - 1/nf2) • When nf > ni then ΔE = (+) • When nf < ni then ΔE = (-)

  20. New Theory Needed to explain more complex atoms! • DeBroglie (1924)-Wave properties of the electron was observed from the diffraction pattern created by a stream of electrons. • Schrodinger (1926)-Developed an equation that correctly accounts for the wave property of the electron and all spectra of atoms. (very complex)

  21. Quantum Theory (current theory of the atom) • Rather than orbits we refer to orbitals. These are 3-dimensional regions of space where there is a high probability of locating the electron. • Heisenberg Uncertainty Principle-it is not possible to know the exact location and momentum (speed) of an electron at the same time. • Quantum Numbers-4 numbers that are used to identify the highest probability location for the electron.

  22. Quantum numbers (reg. chem.) • 1.) Principal Quantum Number (n) • States the main energy level of the electron and also identifies the number of sublevels that are possible. • n=1, n=2, n=3, etc. to n=7 • 2.) Orbital Quantum Number • Identifies the shape of the orbital • s (2 electrons) sphere 1 orbital • P (6 electrons) dumbbell 3 orbitals • d (10 electrons) 4-4 leaf clovers & 1-dumbbell w/doughnut5 orbitals • f (14 electrons) very complex 7 orbitals

  23. Quantum numbers (cont.) • 3.) Magnetic Quantum Number • Identifies the orientation in space (x, y, z) • s  1 orientation • p  3 orientations • d 5 orientations • f 7 orientations • 4.) Spin Quantum Number • States the spin of the electron. • Each orbital can hold at most 2 electrons with opposite spin.

  24. Quantum numbers • 1.) Principal Quantum Number (n) • States the main energy level of the electron and also identifies the number of sublevels that are possible. • n=1, n=2, n=3, etc. to n=7 • 2.) Azimuthal Quantum Number (l) • Values from 0 to n-1 • Identifies the shape of the orbital • l = 0 s sphere 1 orbital • l = 1 p dumbbell 3 orbitals • l = 2 d 4-4 leaf clovers & 1-dumbbell w/doughnut5 orbitals • l = 3 f very complex 7 orbitals

  25. Quantum numbers (cont.) • 3.) Magnetic Quantum Number (ml) • Values from –l  l • States the orientation in space (x, y, z) • ml = 0 s only 1 orientation • ml = -1, 0, +1 p 3 orientations • ml = -2,-1,0,+1,+2 d 5 orientations • ml = -3,-2,-1,0,+1+2,+3 f 7 orientations • 4.) Spin Quantum Number (ms) • Values of +1/2 to -1/2 • States the spin of the electron. • Each orbital can hold at most 2 electrons with opposite spin.

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