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Version 001 – HW08-Wavefunctions – sparks – (52100) This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Mlib 02 0095 001 10.0 points Which of the following provided evidence that the electrons in atoms are arranged in distinct energy levels? 1. the results of the Millikan oil-drop experiment 2. the existence of elements with noninteger values for atomic weights 3. the scattering of α particles by a metal foil 4. the deflection of ions in a mass spectrometer 5. the observation of line spectra from gas discharge tubes correct Explanation: The fact that gases emitted only specific wavelengths of energy suggested that electron energy states are quantized. AtomicSpec 01 002 10.0 points Assume n1 and n2 are two adjacent energy levels of an atom. The emission of radiation with the longest wavelength would occur for which two values of n1 and n2 ? 1. 2,1 7. 3,2 Explanation: The frequency of a photon emitted when an electron moves between levels n1 and n2 is given by the Rydberg equation: 1 1 ν=R − 2 2 n1 n2 A B C 5. 7,6 6. 8,7 correct , Wavefunctions 1s 01 003 10.0 points The graph shows the radial distribution plots for the 1s wavefunctions for H, He, and He+ . 4πr 2 Ψ2 4. 4,3 where R = 3.29 × 1015 Hz. The emission of radiation with the longest wavelength corresponds to that with the smallest frequency. From inspection of the formula above we see that ν is smallest when n1 = 8 and n2 = 7. Conceptual E = hν = Solution: 1 1 hR − gives the energy of the phon21 n22 tons emitted. The emission of radiation with the longest wavelength corresponds to photons with the smallest energy. From the Bohr frequency condition the energy of the emitted photon must be equal to the difference in energy between the higher and lower levels. An energy level diagram for the H-atom shows that as the energy levels get higher, the gaps between them converge; of the transitions listed, the two adjacent levels which are the closest together are n1 = 8 and n2 = 7. Thus a transition from n1 = 8 to n2 = 7 will result in the emission of a photon with the smallest energy, hence the longest wavelength. 2. 6,5 3. 5,4 1 radius Version 001 – HW08-Wavefunctions – sparks – (52100) Which plot is the 1s wavefunction for the He+ ion? 1. C 2. There is no way to know 3. B 4. A correct Explanation: H has one electron and one proton. He has two electrons and two protons. He+ has one electron and two protons. Therefore the electron will have the greatest attraction to the the nuclei with two protons. In He there will be a slight reduction in the effective nuclear charge due to the electron-electron repulsion. In He+ , there is only one electron so the radius will be the smallest. Plot A is peaked closest to the nucleus so it will be the smallest. ChemPrin3e 01 53 54 004 10.0 points How many electrons can have the quantum numbers n = 2; ℓ = 1 in an atom? 1. 4 2 To find the maximum number of electrons that can have the stated values of n and ℓ (and any allowed value of mℓ and ms ), double the number of different values of mℓ . If only n is stated then all allowed values of ℓ, mℓ and ms are possible; each value of ℓ will have its own set of values of mℓ . Total each of these sets of values and then multiply by two to get the number of electrons. If n , ℓ and mℓ are stated then the only other variable is ms which has only two values (only two electrons may share n , ℓ and mℓ ). ChemPrin3e T01 39 005 10.0 points The three quantum numbers for an electron in a hydrogen atom in a certain state are n = 4, ℓ = 2, mℓ = 1. The electron is located in what type of orbital? 1. 4s 2. 3p 3. 4p 4. 4d correct 5. 3d 2. 6 correct Explanation: The notation is nℓ , where n = 1, 2, 3, 4, 5, ..., ℓ = 0, 1, 2, ..., (n − 1) represented as a letter: 3. 3 4. 7 5. 5 Explanation: The notation is nℓ where n = 1, 2, 3, 4, 5, · · · ℓ = 0, 1, 2, · · · , (n − 1) represented by a letter: ℓ value letter 0 s 1 p 2 d 3 f 4 g 5 h mℓ = −ℓ, −(ℓ − 1), −(ℓ − 2), · · · , 0, · · · +(ℓ − 2), +(ℓ − 1), +ℓ 1 and ms = ± . 2 ℓ 0 orbital s 1 2 p d 3 f 4 g 5 etc. h and mℓ = −ℓ, −(ℓ − 1), −(ℓ − 2), ..., 0, ..., +(ℓ − 2), +(ℓ − 1), ℓ. The value of mℓ is not needed to determine the orbital type, as long as it is valid.