HW 05 Atomic Theory-solutions

# HW 05 Atomic Theory-solutions - nanni(arn437 – HW 05...

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Unformatted text preview: nanni (arn437) – HW 05 Atomic Theory – vanden bout – (50985) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The work function for chrominum metal is 4.37 eV. What wavelength of radiation must be used to eject electrons with a velocity of 6000 km / s? Correct answer: 11 . 6266 nm. Explanation: v = 6000 km / s = 6 × 10 6 m / s The wavelength of radiation needed will be the sum of the energy of the work function plus the kinetic energy of the ejected elctron. E work function = (4 . 37 eV) × (1 . 6022 × 10- 19 J / eV) = 7 . 00161 × 10- 19 J E kinetic = 1 2 m v 2 = 1 2 (9 . 10939 × 10- 31 kg) × (6 × 10 6 m / s) 2 = 1 . 63969 × 10- 17 J E total = E work function + E kinetic = 7 . 00161 × 10- 19 J + 1 . 63969 × 10- 17 J = 1 . 70971 × 10- 17 J Since c = ν λ , E = h ν = h c λ λ = h c E = 6 . 626 × 10- 34 m 2 · kg / s 1 . 70971 × 10- 17 J × 3 . × 10 8 m / s = 1 . 16266 × 10- 8 m × 10 9 nm 1 m = 11 . 6266 nm 002 10.0 points Which of the following provided evidence that the electrons in atoms are arranged in distinct energy levels? 1. the existence of elements with noninteger values for atomic weights 2. the deflection of ions in a mass spectrom- eter 3. the observation of line spectra from gas discharge tubes correct 4. the results of the Millikan oil-drop exper- iment 5. the scattering of α particles by a metal foil Explanation: The fact that gases emitted only specific wavelengths of energy suggested that electron energy states are quantized. 003 10.0 points Assume n 1 and n 2 are two adjacent energy levels of an atom. The emission of radiation with the longest wavelength would occur for which two values of n 1 and n 2 ? 1. 6,5 2. 5,4 3. 2,1 4. 8,7 correct 5. 7,6 6. 3,2 7. 4,3 Explanation: The frequency of a photon emitted when an electron moves between levels n 1 and n 2 is given by the Rydberg equation: ν = R parenleftbigg 1 n 2 1- 1 n 2 2 parenrightbigg , nanni (arn437) – HW 05 Atomic Theory – vanden bout – (50985) 2 where R = 3 . 29 × 10 15 Hz. The emission of radiation with the longest wavelength corre- sponds to that with the smallest frequency. From inspection of the formula above we see that ν is smallest when n 1 = 8 and n 2 = 7. Conceptual Solution : E = h ν = h R parenleftbigg 1 n 2 1- 1 n 2 2 parenrightbigg gives the energy of the pho- tons emitted. The emission of radiation with the longest wavelength corresponds to pho- tons with the smallest energy. From the Bohr frequency condition the energy of the emit- ted photon must be equal to the difference in energy between the higher and lower lev- els. An energy level diagram for the H-atom shows that as the energy levels get higher, the gaps between them converge; of the transi- tions listed, the two adjacent levels which are the closest together are n 1 = 8 and n 2 = 7....
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HW 05 Atomic Theory-solutions - nanni(arn437 – HW 05...

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