Acceleration/Radiation Spectral Lines

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Presentation transcript:

Acceleration/Radiation Spectral Lines Problems with the Rutherford Atom Acceleration/Radiation Spectral Lines

Spectral lines Rydberg’s Formula: (FYI) Energy from excited atoms demo Rydberg’s Formula: (FYI) 1/ = R(1/22 - 1/n2), n = 3, 4, ...(Balmer) (Visible) 1/ = R(1/12 - 1/n2), n = 2, 3, ...(Lyman) (UV) 1/ = R(1/32 - 1/n2), n = 4, 5, ...(Paschen) (IR) (R = 1.097 x 10-7 m-1) H He Sun

Bohr’s Quantum Atom Only certain orbits are allowed “stationary states” Electron transitions create photons 1/ = R(1/22 - 1/n2), n = 3, 4, ...(Balmer) (Visible) 1/ = R(1/12 - 1/n2), n = 2, 3, ...(Lyman) (UV) 1/ = R(1/32 - 1/n2), n = 4, 5, ...(Paschen) (IR)

Example: What is the wavelength of the first Lyman line? (122 nm) The first Lyman line is a transition from -3.4 eV to -13.6 eV, so it releases 10.2 eV of energy. A photon with this energy has this wavelength: E = (10.2)(1.602E-19) = 1.63404E-18 J E = hc/λ, λ = hc/E = (6.626E-34)(3.00E8)/(1.63404E-18) = 1.21649E-07 m =

Limitations of Bohr’s model Works well for H, but doesn’t even work for He Did not explain Spectral fine structure Brightness of lines Molecular bonds Theory was not complete. But otherwise it generally kicked tuckus

1. What possible photon energies can you get from these energy levels 1. What possible photon energies can you get from these energy levels? (there are 6 different ones) 1 4 9 3 8 5 -5.0 eV -6.0 eV -9.0 eV -14.0 eV 1, 4, 9, 3, 8, and 5 eV

2. What is the wavelength of the photon released from the third Lyman spectral line (from -.85 to -13.6 eV)? E = hf = hc/  E = -.85 - -13.6 = 12.75 eV E = (12.75 eV)(1.602E-19J/eV) = 2.04E-18J  = hc/E = 97.3 = 97 nm 97 nm

3. What is the wavelength of the photon released from the second Balmer spectral line (from –0.85 to -3.4 eV)? E = hf = hc/  E = -0.85 -3.4 = 2.55 eV E = (2.55 eV)(1.602E-19J/eV) = 4.09E-19J  = hc/E = 487 = 490 nm 490 nm

4. An 102.5 nm photon is emitted. What is the energy of this photon in eV, and what transition occurred? E = hf = hc/  (6.626E-34)(3.00E8)/(86.4E-9) = 2.30069E-18 J (2.30069E-18 J)/(1.602E-19) = 12.1 eV This could be the second Lyman line 12.1 eV