Orbital shapes and bohr's theory
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Publicado: 6 de septiembre de 2012
Lecture notes: CHEM103 Spring 2007
9/3/12 8:05 PM
Lecture notes: CHEM103 Spring 2008 – October 2, 2008
OBJECTIVES FOR TODAY: 0) REVIEW: quantized electron orbits 1) DeBroglie and wave-particleduality of all matter a. …including electrons – this explains the quantized orbits 2) emission and absorption of light by atoms: INTERACTION OF LIGHT W/ MATTER a. calculating electron energy changesin atomic transitions using Planck AND Rydberg equations b. absorption of energy c. emission of energy d. ground state e. ionization energy 3) electrons as a 3-D wave a. uncertainty principle:probability NOT location! b. Schrodinger’s equation – yikes! c. orbital shapes (waveforms) of electrons d. describing them with 4 quantum numbers
THE PROBLEM: electrons in orbits around a central nucleusinteractions between charges should cause the electron to spiral into the nucleus NOT GOOD!!! THE SOLUTION: just like light, the orbital velocity or radius (energy) of these electron orbits areQUANTIZED! (ONLY CERTAIN SOLUTIONS TO THE PROBLEM ALLOWED!) BUT THIS ONLY WORKS IF ELECTRONS ARE A WAVE, JUST LIKE LIGHT
http://www.uwec.edu/boulteje/Boulter103Notes/2October.htm
Page 1 of 10 Lecture notes: CHEM103 Spring 2007
9/3/12 8:05 PM
This is first theorized by Prince Louis-Victor Pierre Raymond de Broglie -- the wave nature of all matter! For ALL objects – we are all quantized!!!REVIEW: Einstein: E = mc2 Planck: E = hc/l l = h/mv ! The electron has a wavelength! (Yep – another Nobel prize!)
DeBroglie:
(mass must be in kg because “h” has units of J*s, and 1 J = kgm2/s2)Example: a baseball has a wavelength determined by… 5.00 oz @ 100.0 mph mb = 5 oz (16 oz/lb; 453.6 g/lb) = 0.142 kg vb = 100 mi/hr (5280 ft/mi; 12 in/ft; 2.54 cm/in; 3600 s/hr) = 44.70 m/sCalculate: wavelength of an electron h = 6.626 x 10-34 J·s me = 9.109 x 10-31 kg ve = 2.19 x 106 m/sec = 3.32 x 10-10 m
http://www.uwec.edu/boulteje/Boulter103Notes/2October.htm
Page 2 of 10...
9/3/12 8:05 PM
Lecture notes: CHEM103 Spring 2008 – October 2, 2008
OBJECTIVES FOR TODAY: 0) REVIEW: quantized electron orbits 1) DeBroglie and wave-particleduality of all matter a. …including electrons – this explains the quantized orbits 2) emission and absorption of light by atoms: INTERACTION OF LIGHT W/ MATTER a. calculating electron energy changesin atomic transitions using Planck AND Rydberg equations b. absorption of energy c. emission of energy d. ground state e. ionization energy 3) electrons as a 3-D wave a. uncertainty principle:probability NOT location! b. Schrodinger’s equation – yikes! c. orbital shapes (waveforms) of electrons d. describing them with 4 quantum numbers
THE PROBLEM: electrons in orbits around a central nucleusinteractions between charges should cause the electron to spiral into the nucleus NOT GOOD!!! THE SOLUTION: just like light, the orbital velocity or radius (energy) of these electron orbits areQUANTIZED! (ONLY CERTAIN SOLUTIONS TO THE PROBLEM ALLOWED!) BUT THIS ONLY WORKS IF ELECTRONS ARE A WAVE, JUST LIKE LIGHT
http://www.uwec.edu/boulteje/Boulter103Notes/2October.htm
Page 1 of 10 Lecture notes: CHEM103 Spring 2007
9/3/12 8:05 PM
This is first theorized by Prince Louis-Victor Pierre Raymond de Broglie -- the wave nature of all matter! For ALL objects – we are all quantized!!!REVIEW: Einstein: E = mc2 Planck: E = hc/l l = h/mv ! The electron has a wavelength! (Yep – another Nobel prize!)
DeBroglie:
(mass must be in kg because “h” has units of J*s, and 1 J = kgm2/s2)Example: a baseball has a wavelength determined by… 5.00 oz @ 100.0 mph mb = 5 oz (16 oz/lb; 453.6 g/lb) = 0.142 kg vb = 100 mi/hr (5280 ft/mi; 12 in/ft; 2.54 cm/in; 3600 s/hr) = 44.70 m/sCalculate: wavelength of an electron h = 6.626 x 10-34 J·s me = 9.109 x 10-31 kg ve = 2.19 x 106 m/sec = 3.32 x 10-10 m
http://www.uwec.edu/boulteje/Boulter103Notes/2October.htm
Page 2 of 10...
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