140 likes | 447 Views
Niels Bohr and the quantum atom Contents: Problems in nucleus land Spectral lines and Rydberg’s formula Photon wavelengths from transition energies Electron in a box Schr ö dinger Limitations of Bohr’s model. Niels Bohr 1881 - 1962. Problems with the Rutherford Atom.
E N D
Niels Bohr and the quantum atom • Contents: • Problems in nucleus land • Spectral lines and Rydberg’s formula • Photon wavelengths from transition energies • Electron in a box • Schrödinger • Limitations of Bohr’s model Niels Bohr 1881 - 1962
Problems with the Rutherford Atom • Acceleration/Radiation • Spectral Lines TOC
Spectral lines • 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 TOC
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? 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 = 122 nm
Whiteboards: Bohr Photons 1 | 2 | 3 | 4 TOC
9 5 1 4 3 8 What possible photon energies can you get from these energy levels? (there are 6 different ones) -5.0 eV -6.0 eV -9.0 eV -14.0 eV 1, 4, 9, 3, 8, and 5 eV
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 W 97 nm
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 W 490 nm
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 W 12.1 eV
“Electron in a box” • Why are only certain orbits allowed? (Demo) TOC
Try this: What would be the kinetic energy of an electron in the ground state of a “box” that is about 0.1 nm in length? me = 9.11x10-31 kg Ek = 12(6.626E-34)2/(8(9.11E-31)(.1E-9)2) = 6.02413E-18 J (≈ 38 eV) W 6 x 10-18 J
Schrödinger and the quantum atom Schrödinger solves for hydrogen atom The electron is represented by a wave Can only be solved for H, singly ionised He TOC
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 TOC