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Arrangement of the Electrons Chapter 4 (reg.)

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Presentation on theme: "Arrangement of the Electrons Chapter 4 (reg.)"— Presentation transcript:

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 RESULT-An electromagnetic wave.
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. 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. Click

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 formula v= c λ

8 Frequency = 4.7x1014 Hz (cycles per second)
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 (cycles per second)

9 B. EM Spectrum  = ?  = c   = 434 nm = 4.34  10-7 m
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 WORK:  = c  = 3.00  108 m/s 4.34  10-7 m  = 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 between Energy and ν
E = h∙ν E = energy (joule) h = Planck’s constant = 6.63 x j∙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 lines for 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 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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