An atomic emission spectrum of hydrogen shows three wavelengths: 656.5 nm, 486.3 nm, and 434.2 nm. Assign these wavelengths to transitions in the hydrogen atom. Match the items in the left column to the appropriate blanks in the sentences on the right. Reset Help 1 1 AE 1.1 x 107 2 nimal nitial the wavelengths of the visual The wavelengths of the three transitions are region. Therefore, it can be assumed that the electrons return to the level. shorter than longer than n = 7 The difference in energy between the two levels initial and nfinal is given by the 1 1 equation AE-2.18 x 10-18 2 ninal After calculating the energy difference using the given wavelengths, solving the equation for initial, and subtituting the proper values the following initial energy levels can be 1 1 AE -2.18 x 10-18 2 ninal ninitial obtained: n = 1 for 434.2nm. 1 AE -2.18 x 10-18 initial ninal n = 8 equal to n=6 n = 2 n = 3 n = 4 n = 5 Submit Request Answer for 656.5nm, for 486.3nm, and

An atomic emission spectrum of hydrogen shows three wavelengths: 656.5 nm, 486.3 nm, and 434.2 nm. Assign these wavelengths to transitions in the hydrogen atom. Match the items in the left column to the appropriate blanks in the sentences on the right. Reset Help 1 1 AE 1.1 x 107 2 nimal nitial the wavelengths of the visual The wavelengths of the three transitions are region. Therefore, it can be assumed that the electrons return to the level. shorter than longer than n = 7 The difference in energy between the two levels initial and nfinal is given by the 1 1 equation AE-2.18 x 10-18 2 ninal After calculating the energy difference using the given wavelengths, solving the equation for initial, and subtituting the proper values the following initial energy levels can be 1 1 AE -2.18 x 10-18 2 ninal ninitial obtained: n = 1 for 434.2nm. 1 AE -2.18 x 10-18 initial ninal n = 8 equal to n=6 n = 2 n = 3 n = 4 n = 5 Submit Request Answer for 656.5nm, for 486.3nm, and

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Short Answer 1. Explain the historical development of the atomic model from Democritus to Bohr. 2. What is the difference between an emission spectrum and an absorption spectrum for a specific element? 3. What does it mean to say something is radioactive?
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The K series of the discrete spectrum of tungsten contains wavelengths of 0.0185 nm, 0.0209 nm, and 0.0215 nm. The K-shell ionization energy is 69.5 keV. Determine the ionization energies of the L, M, and N shells. Step 1 Characteristic x-rays of the K series are produced when an electron in an upper energy level of a many- electron atom drops down to fill a vacancy in the K (n = 1) shell. Assuming that the given wavelengths are the three longest wavelengths in the K series of tungsten, the transitions that produce them are from the N (n 4) shell to the K shell, from the M (n = 3) shell to the K shell, and from the L (n = 2) shell to the K shell, respectively. These transitions are shown in the energy level diagram, where 1, = 0.0185 nm, = 0.0209 nm, and 13 12 EK = 0.0215 nm. If the ionization energy of the K shell is 69.5 keV, then = -69.5 keV. En N shell EM M shell EL L shell Ex- K shell The photon produced in a transition from an initial state to a final state has energy hc Ephoton…
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