Physics for Scientists and Engineers with Modern Physics
4th Edition
ISBN: 9780131495081
Author: Douglas C. Giancoli
Publisher: Addison-Wesley
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Question
Chapter 37, Problem 21Q
To determine
The reason why absorption lines corresponding to only the Lyman series are observed.
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Students have asked these similar questions
When a wide spectrum of light passes through hydrogengas at room temperature, absorption lines are observedthat correspond only to the Lyman series. Why don’t weobserve the other series?
The light observed that is emitted by a hydrogen atom is explained by a simple model of its structure with one proton in its nucleus and an electron bound to it, but only with internal energies of the atom satisfying
EH=−RH/n2EH=−RH/n2
where RHRH is the Rydberg constant and nn is an integer such as 1, 2, 3 ... and so on. When a hydrogen atom in an excited state emits light, the photon carries away energy and the atom goes into a lower energy state.
Be careful about units. The Rydberg constant in eV is
13.605693009 eV
That would be multiplied by the charge on the electron 1.602× 10-19 C to give
2.18× 10-18 J
A photon with this energy would have a frequency f such that E=hf. Its wavelength would be λ = c/f = hc/E. Sometimes it is handy to measure the Rydberg constant in units of 1/length for this reason. You may see it given as 109737 cm-1 if you search the web, so be aware that's not joules.
The following questions are intended to help you understand the connection between…
The wavelengths of the Lyman series for hydrogen are given by = RH(1-2), = 2, 3, 4, ...
1/2
(a) Calculate the wavelengths of the first three lines in this series.
nm
nm
nm
(b) Identify the region of the electromagnetic spectrum in which these lines appear.
O ultraviolet region
O infrared region
O x-ray region
O visible light region
O gamma ray region
Chapter 37 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 37.2 - Prob. 1AECh. 37.2 - Prob. 1BECh. 37.4 - Prob. 1CECh. 37.7 - Prob. 1DECh. 37.7 - Prob. 1EECh. 37.11 - Prob. 1FECh. 37 - Prob. 1QCh. 37 - Prob. 2QCh. 37 - Prob. 3QCh. 37 - Prob. 4Q
Ch. 37 - Prob. 5QCh. 37 - Prob. 6QCh. 37 - Prob. 7QCh. 37 - Prob. 8QCh. 37 - Prob. 9QCh. 37 - Prob. 10QCh. 37 - Prob. 11QCh. 37 - Prob. 12QCh. 37 - Prob. 13QCh. 37 - Prob. 14QCh. 37 - Prob. 15QCh. 37 - Prob. 16QCh. 37 - Prob. 17QCh. 37 - Prob. 18QCh. 37 - Prob. 19QCh. 37 - Prob. 20QCh. 37 - Prob. 21QCh. 37 - Prob. 22QCh. 37 - Prob. 23QCh. 37 - Prob. 24QCh. 37 - Prob. 25QCh. 37 - Prob. 26QCh. 37 - Prob. 27QCh. 37 - Prob. 28QCh. 37 - Prob. 1PCh. 37 - Prob. 2PCh. 37 - Prob. 3PCh. 37 - Prob. 4PCh. 37 - Prob. 5PCh. 37 - Prob. 6PCh. 37 - Prob. 7PCh. 37 - Prob. 8PCh. 37 - Prob. 9PCh. 37 - Prob. 10PCh. 37 - Prob. 11PCh. 37 - Prob. 12PCh. 37 - Prob. 13PCh. 37 - Prob. 14PCh. 37 - Prob. 15PCh. 37 - Prob. 16PCh. 37 - Prob. 17PCh. 37 - Prob. 18PCh. 37 - Prob. 19PCh. 37 - Prob. 20PCh. 37 - Prob. 21PCh. 37 - Prob. 22PCh. 37 - Prob. 23PCh. 37 - Prob. 24PCh. 37 - Prob. 25PCh. 37 - Prob. 26PCh. 37 - Prob. 27PCh. 37 - Prob. 28PCh. 37 - Prob. 29PCh. 37 - Prob. 30PCh. 37 - Prob. 31PCh. 37 - Prob. 32PCh. 37 - Prob. 33PCh. 37 - Prob. 34PCh. 37 - Prob. 35PCh. 37 - Prob. 36PCh. 37 - Prob. 37PCh. 37 - Prob. 38PCh. 37 - Prob. 39PCh. 37 - Prob. 40PCh. 37 - Prob. 41PCh. 37 - Prob. 42PCh. 37 - Prob. 43PCh. 37 - Prob. 44PCh. 37 - Prob. 45PCh. 37 - Prob. 46PCh. 37 - Prob. 47PCh. 37 - Prob. 48PCh. 37 - Prob. 49PCh. 37 - Prob. 50PCh. 37 - Prob. 51PCh. 37 - Prob. 52PCh. 37 - Prob. 53PCh. 37 - Prob. 54PCh. 37 - Prob. 55PCh. 37 - Prob. 56PCh. 37 - Prob. 57PCh. 37 - Prob. 58PCh. 37 - Prob. 59PCh. 37 - Prob. 60PCh. 37 - Prob. 61PCh. 37 - Prob. 62PCh. 37 - Prob. 63PCh. 37 - Prob. 64PCh. 37 - Prob. 65PCh. 37 - Prob. 66PCh. 37 - Prob. 67PCh. 37 - Prob. 68PCh. 37 - Prob. 69PCh. 37 - Prob. 70PCh. 37 - Prob. 71PCh. 37 - Prob. 72GPCh. 37 - Prob. 73GPCh. 37 - Prob. 74GPCh. 37 - Prob. 75GPCh. 37 - Prob. 76GPCh. 37 - Prob. 77GPCh. 37 - Prob. 78GPCh. 37 - Prob. 79GPCh. 37 - Prob. 80GPCh. 37 - Prob. 81GPCh. 37 - Prob. 82GPCh. 37 - Prob. 83GPCh. 37 - Prob. 84GPCh. 37 - Prob. 85GPCh. 37 - Prob. 86GPCh. 37 - Prob. 87GPCh. 37 - Prob. 88GPCh. 37 - Prob. 89GPCh. 37 - Prob. 90GPCh. 37 - Prob. 91GPCh. 37 - Prob. 92GPCh. 37 - Prob. 93GPCh. 37 - Show that the wavelength of a particle of mass m...Ch. 37 - Prob. 95GPCh. 37 - Prob. 96GPCh. 37 - Prob. 97GPCh. 37 - Prob. 98GPCh. 37 - Prob. 99GPCh. 37 - Prob. 100GP
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- The Balmer series for hydrogen was discovered before either the Lyman or the Paschen series. Why?arrow_forwardDo the Balmer series and the Lyman series overlap? Why? Why not? (Hint: calculate the shortest Balmer line and the longest Lyman line.)arrow_forwardWhen a hydrogen atom is in its ground state, what are the shortest and longest wavelengths of the photons it can absorb without being ionized?arrow_forward
- In extreme-temperature environments, such as those existing in a solar corona, atoms may be ionized by undergoing collisions with other atoms. One example of such ionization in the solar corona is the presence of C5+ ions, detected in the Fraunhofer spectrum. (a) By what factor do the energies of the C5+ ion scale compare to the energy spectrum of a hydrogen atom? (b) What is the wavelength of the first line in the Paschen series of C5+ ? (c) In what part of the spectrum are these lines located?arrow_forwardWhy are X-rays emitted only for electron transitions to inner shells? What type of photon is emitted for transitions between outer shells?arrow_forwardWhat is the longest - wavelength line in nanometers in the infrared series for hydrogen where m = 3?arrow_forward
- The wavelengths of the Lyman series for hydrogen are given by 1 λ = RH 1 − 1 n2 ,n = 2, 3, 4, . . . (a) Calculate the wavelengths of the first three lines in this series. nm nm nm (b) Identify the region of the electromagnetic spectrum in which these lines appear. infrared regionvisible light region x-ray regionultraviolet regiongamma ray regionarrow_forwardYou measure a wavelength of 397.1 nm in a spectroscopy experiment. You identify this as a particular transition from the hydrogen Balmer series. Which transition is it?arrow_forwarda) Calculate the energy of the emissive transition with the lowest energy possible for the Lyman series, for a mole of hydrogen atoms. Express your answer in joules/mol. b) Is this transition in the visible spectral domain? If not, in which region is it located?arrow_forward
- The electron, in a hydrogen atom, is in its second excited state. Calculate the wavelength of the lines in the Lyman series, that can be emitted through the permissible transitions of this electron. (Given the value of Rydberg constant, R = 1.1 × 107 m-1 )arrow_forward3:09 O O O 63° A X • N N O 5G „ll Quizzes a (absorption) Brackett series Paschen series Lyman series (emission) Balmer series Paschen series (emission) n= 2 n=3 n=4 .... Lyman series n-5 (a) (b) e These pictures refer to the energy levels of a hydrogen atom. You can find the error in both parts, (a) and (b). The arrows labeled "emission" in (a), and all the arrows in (b), indicate a transition in which an electron jumps from a higher- energy state to a lower-energy state. The different "series" of emission lines are characterized by the index n of the low- energy state in which the electron ends up. In particular, the Lyman series consists of all transitions that end up in the n=1 energy level, with an initial energy level that corresponds to the label n = 2, 3, 4, 5, etc. One of these values of n is not shown as an arrow in the Lyman emission series in figures (a) or (b). This is a significant error because that particular spectral line is very important in astronomy. Pick the value…arrow_forwardIn hydrogen’s characteristic spectra, each series (Lyman, Balmer, etc) has a “series limit”, where the wavelengths at one end of the series tend to “bunch up”, approaching a single limiting value. part a: Is it at the short-wavelength or the long-wavelength end of the series that this series limit occurs? part b: What is it about hydrogen’s allowed energies that leads to this phenomenon?arrow_forward
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