Physics for Scientists and Engineers with Modern Physics
4th Edition
ISBN: 9780131495081
Author: Douglas C. Giancoli
Publisher: Addison-Wesley
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Chapter 37, Problem 21Q
To determine
The reason why absorption lines corresponding to only the Lyman series are observed.
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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…
Calculate the frequency of the n = 4 line in the Lyman series of hydrogen. v゠
(Please type answer no write by hend)
a) 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?
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
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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_forwardThe 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_forward
- In 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_forwardThe wavelengths of the Lyman series for hydrogen are given by (a) Calculate the wavelengths of the first three lines in this series. (b) Identify the region of the electromagnetic spectrum in which these lines appeararrow_forwardThe shortest wavelength of the Lyman series of hydrogen is 91.13 nm. Find the three longest wavelengths in this series! Please Answer my question Im needed Max 30 minutes please.... Thank uarrow_forward
- It is possible that a muon be captured by a proton to form a muonic atom. A muon is identic to an electron, except when your mass, which is m = 105.7 MeV/c^2. What ia the smallest wave length for a Lyman series for this atom? Give your answer in pm.arrow_forward(a) The Lyman series in hydrogen is the transition from energy levels n = 2, 3, 4, ... to the ground state n = 1. The energy levels are given by 13.60 eV En n- (i) What is the second longest wavelength in nm of the Lyman series? (ii) What is the series limit of the Lyman series? [1 eV = 1.602 x 1019 J, h = 6.626 × 10-34 J.s, c = 3 × 10° m.s] %3D Two emission lines have wavelengts A and + A2, respectively, where AA <<2. Show that the angular separation A0 in a grating spectrometer is given aproximately by (b) A0 = V(d/m)-2 where d is the grating constant and m is the order at which the lines are observed.arrow_forwardThe Lyman series comprises a set of spectral lines. All of these lines involve a hydrogen atom whose electron undergoes a change in energy level, either beginning at the n = 1 level (in the case of an absorption line) or ending there (an emission line). The inverse wavelengths for the Lyman series in hydrogen are given by 1 - where n = 2, 3, 4, ... and the Rydberg constant R, = 1.097 x 10' m-. (Round your answers to at least one decimal place. Enter your answers in nm.) %3D (a) Compute the wavelength for the first line in this series (the line corresponding to n = 2). nm (b) Compute the wavelength for the second line in this series (the line corresponding to n = 3). nm (c) Compute the wavelength for the third line in this series (the line corresponding to n = 4). nm (d) In which part of the electromagnetic spectrum do these three lines reside? O x-ray region O ultraviolet region O infrared region O gamma ray region O visible light regionarrow_forward
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