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ninitial = 3 ninitial = 4 %3D Ninitial = 5 Ninitial = 6 %3D

ninitial = 3 ninitial = 4 %3D Ninitial = 5 Ninitial = 6 %3D

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter5: Electron Configurations And The Periodic Table
Section: Chapter Questions
Problem 114QRT
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Hello! I needed help with this part of my chemistry lab (#2). The first picture is the instructions in the the theory. And the second picture is part of the results/calculations. Thank you!
A few of the energy levels of the hydrogen atom are shown below (given in units of kJ/mol of photons): n = oc OkJ n = 7 n = 6 n = 5 Pfund n = 4 -82 kJ Brackett n = 3 -146 kJ Paschen n = 2 -328 kJ Balmer n= 1 -1312 kJ Lyman These energies can also be reported in units of J/photon as show here: Principle Quantum Number (n): n = c0 (ionization occurs) n= 3 (2nd excited state) n=2 (1st excited state) n = 1 (ground state) Energy: -2.421 X 10-19 J -5.448 X 10-19 J -2.179 X 10-18 J The energy of the photon emitted when the electron in the hydrogen atom makes a transition from the first excited state (n=2) to the ground state (n=1) is: Ephoton = (-5.448 x 10-19 J) - (-2.179 x 10-18 J) = 1.634 X 10-18 J hc Rearranging Plank's equation, 1 = , we can calculate the wavelength of this transition's E photon (6.626x10-34 J . s)× (2.997×10* ") photon: 1 = S =1.216×10m 1.634x10 18 J This electron transition corresponds to a wavelength of 1.216 x 10" m, which is 121.6 nm. This electronic transition leads to a wavelength that is higher energy than the range which we perceive as visible light. This transition can't see it, it is real and an instrument (UV spectrometer) can. ices a photon of energy in the ultraviolet region. While we In this experiment, we will be observing and recording the four visible transitions' wavelengths for hydrogen. Then we will use the Rydberg equation to determine what the theoretical wavelengths are. These will then be compared with a percent error. For the other elements, we will draw colored lines in to reflect qualitatively the fingerprint. 1 AE = RH n final (Rydberg Equation, where R, =2.18x10 J) ninitial
Transcribed Image Text:A few of the energy levels of the hydrogen atom are shown below (given in units of kJ/mol of photons): n = oc OkJ n = 7 n = 6 n = 5 Pfund n = 4 -82 kJ Brackett n = 3 -146 kJ Paschen n = 2 -328 kJ Balmer n= 1 -1312 kJ Lyman These energies can also be reported in units of J/photon as show here: Principle Quantum Number (n): n = c0 (ionization occurs) n= 3 (2nd excited state) n=2 (1st excited state) n = 1 (ground state) Energy: -2.421 X 10-19 J -5.448 X 10-19 J -2.179 X 10-18 J The energy of the photon emitted when the electron in the hydrogen atom makes a transition from the first excited state (n=2) to the ground state (n=1) is: Ephoton = (-5.448 x 10-19 J) - (-2.179 x 10-18 J) = 1.634 X 10-18 J hc Rearranging Plank's equation, 1 = , we can calculate the wavelength of this transition's E photon (6.626x10-34 J . s)× (2.997×10* ") photon: 1 = S =1.216×10m 1.634x10 18 J This electron transition corresponds to a wavelength of 1.216 x 10" m, which is 121.6 nm. This electronic transition leads to a wavelength that is higher energy than the range which we perceive as visible light. This transition can't see it, it is real and an instrument (UV spectrometer) can. ices a photon of energy in the ultraviolet region. While we In this experiment, we will be observing and recording the four visible transitions' wavelengths for hydrogen. Then we will use the Rydberg equation to determine what the theoretical wavelengths are. These will then be compared with a percent error. For the other elements, we will draw colored lines in to reflect qualitatively the fingerprint. 1 AE = RH n final (Rydberg Equation, where R, =2.18x10 J) ninitial
Results/Calculations: 1. For each of the observed wavelengths, calculated the energy of the photon using equation Keep track of units. Show work. -34 -28 3.035X10J Red 6SS nm = E= hc 6.626x10.300x 10 Turquoise 485 nm - E = hc = b.b26x10"3.00x10 = 4.0o9 x10?J 485 nm %3D Violet 435 nm E= hc blozlox 10"3.00x 108_ 4570 x1028 J 435 nm Purple 410 nm %3D = 4,848 x10-28J 6. 2 410nm 2. The initial energy level for each of the observed lines, ninitial, is given below. The lines observed in this lab are transitions between these initial levels and a final level of 2 (ninal = 2). Use this information to calculate the energy of each transition (in units of J per photon) Look carefully for which equation you should be using. ninitial = 3 Ninitial = 4 Ninitial = 5 Ninitial = 6
Transcribed Image Text:Results/Calculations: 1. For each of the observed wavelengths, calculated the energy of the photon using equation Keep track of units. Show work. -34 -28 3.035X10J Red 6SS nm = E= hc 6.626x10.300x 10 Turquoise 485 nm - E = hc = b.b26x10"3.00x10 = 4.0o9 x10?J 485 nm %3D Violet 435 nm E= hc blozlox 10"3.00x 108_ 4570 x1028 J 435 nm Purple 410 nm %3D = 4,848 x10-28J 6. 2 410nm 2. The initial energy level for each of the observed lines, ninitial, is given below. The lines observed in this lab are transitions between these initial levels and a final level of 2 (ninal = 2). Use this information to calculate the energy of each transition (in units of J per photon) Look carefully for which equation you should be using. ninitial = 3 Ninitial = 4 Ninitial = 5 Ninitial = 6
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