PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 27, Problem 17CQ
To determine
The reason why every line in the absorption spectrum of hydrogen is present in the emission spectrum, but not every line in the emission spectrum is present in the absorption spectrum.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The wavelengths of the Lyman series for hydrogen are given by:
= RH(1-1), n = 2, 3, 4, ...
For the second of this series; calculate the energy (in eV).
Note: 1.60 x 10^-19 J = 1.0 eV
O 4.10 x 10^3 eV
2.12 x 10^3 eV
3²
O 1.21 x 10^3 eV
3.30 x 10^3 eV
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
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 region
Chapter 27 Solutions
PHYSICS
Ch. 27.2 - Prob. 27.2CPCh. 27.3 - Prob. 27.1PPCh. 27.3 - Prob. 27.2PPCh. 27.3 - Prob. 27.3CPCh. 27.3 - Prob. 27.3PPCh. 27.4 - Prob. 27.4PPCh. 27.5 - Prob. 27.5CPCh. 27.5 - Prob. 27.5PPCh. 27.7 - Prob. 27.7CPCh. 27.7 - Prob. 27.6PP
Ch. 27.7 - Prob. 27.7PPCh. 27.7 - Prob. 27.8PPCh. 27.8 - Prob. 27.9PPCh. 27 - Prob. 1CQCh. 27 - Prob. 2CQCh. 27 - Prob. 3CQCh. 27 - Prob. 4CQCh. 27 - Prob. 5CQCh. 27 - Prob. 6CQCh. 27 - Prob. 7CQCh. 27 - Prob. 8CQCh. 27 - Prob. 9CQCh. 27 - Prob. 10CQCh. 27 - Prob. 11CQCh. 27 - Prob. 12CQCh. 27 - Prob. 13CQCh. 27 - Prob. 14CQCh. 27 - Prob. 15CQCh. 27 - Prob. 16CQCh. 27 - Prob. 17CQCh. 27 - Prob. 18CQCh. 27 - Prob. 19CQCh. 27 - Prob. 20CQCh. 27 - Prob. 21CQCh. 27 - Prob. 22CQCh. 27 - Prob. 23CQCh. 27 - Prob. 1MCQCh. 27 - Prob. 2MCQCh. 27 - Prob. 3MCQCh. 27 - Prob. 4MCQCh. 27 - Prob. 5MCQCh. 27 - Prob. 6MCQCh. 27 - Prob. 7MCQCh. 27 - Prob. 8MCQCh. 27 - Prob. 9MCQCh. 27 - Prob. 10MCQCh. 27 - Prob. 1PCh. 27 - Prob. 2PCh. 27 - Prob. 3PCh. 27 - Prob. 4PCh. 27 - Prob. 5PCh. 27 - Prob. 6PCh. 27 - Prob. 7PCh. 27 - Prob. 8PCh. 27 - Prob. 9PCh. 27 - Prob. 10PCh. 27 - Prob. 11PCh. 27 - Prob. 12PCh. 27 - Prob. 13PCh. 27 - Prob. 14PCh. 27 - Prob. 15PCh. 27 - Prob. 16PCh. 27 - Prob. 17PCh. 27 - Prob. 18PCh. 27 - Prob. 19PCh. 27 - Prob. 20PCh. 27 - Prob. 21PCh. 27 - Prob. 22PCh. 27 - Prob. 23PCh. 27 - Prob. 24PCh. 27 - Prob. 25PCh. 27 - Prob. 26PCh. 27 - Prob. 27PCh. 27 - Prob. 28PCh. 27 - Prob. 29PCh. 27 - Prob. 30PCh. 27 - Prob. 31PCh. 27 - Prob. 32PCh. 27 - Prob. 33PCh. 27 - Prob. 34PCh. 27 - Prob. 35PCh. 27 - Prob. 36PCh. 27 - Prob. 37PCh. 27 - Prob. 38PCh. 27 - Prob. 39PCh. 27 - Prob. 40PCh. 27 - Prob. 41PCh. 27 - Prob. 42PCh. 27 - Prob. 43PCh. 27 - Prob. 44PCh. 27 - Prob. 45PCh. 27 - Prob. 46PCh. 27 - Prob. 47PCh. 27 - Prob. 48PCh. 27 - Prob. 49PCh. 27 - Prob. 50PCh. 27 - Prob. 51PCh. 27 - Prob. 52PCh. 27 - Prob. 53PCh. 27 - Prob. 54PCh. 27 - Prob. 55PCh. 27 - Prob. 56PCh. 27 - Prob. 57PCh. 27 - Prob. 58PCh. 27 - Prob. 59PCh. 27 - Prob. 61PCh. 27 - Prob. 60PCh. 27 - Prob. 62PCh. 27 - Prob. 64PCh. 27 - Prob. 63PCh. 27 - Prob. 66PCh. 27 - Prob. 65PCh. 27 - Prob. 68PCh. 27 - Prob. 67PCh. 27 - Prob. 70PCh. 27 - Prob. 69PCh. 27 - Prob. 72PCh. 27 - Prob. 71PCh. 27 - Prob. 74PCh. 27 - Prob. 73PCh. 27 - Prob. 75PCh. 27 - Prob. 76PCh. 27 - Prob. 77PCh. 27 - Prob. 78PCh. 27 - Prob. 79PCh. 27 - Prob. 80PCh. 27 - Prob. 82PCh. 27 - Prob. 81PCh. 27 - Prob. 84PCh. 27 - Prob. 83PCh. 27 - Prob. 86PCh. 27 - Prob. 85PCh. 27 - Prob. 88PCh. 27 - Prob. 87PCh. 27 - Prob. 89PCh. 27 - Prob. 90PCh. 27 - Prob. 91PCh. 27 - Prob. 92PCh. 27 - Prob. 93PCh. 27 - Prob. 94PCh. 27 - Prob. 95PCh. 27 - Prob. 96P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- If an atom has an election in the n = 5 state with m = 3, what are the possible values of l?arrow_forwardConstruct the energy-level diagram for the He+ ion(like Fig. 27–29).arrow_forwardThe radial probability density of a hydrogen wavefunction in the 1s state is given by P(r) = |4rr2 (R13 (r))²| and the radial wavefunction R1s (r) = a0 , where ao is 3/2 the Bohr radius. Using the standard integral x"e - ka dx n! calculate the standard deviation in the radial position from the nucleus for the 1s state in the Hydrogen atom. Give your answer in units of the Bohr radius ao.arrow_forward
- 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…arrow_forwardAn energy of about 21 eV is required to excite an electron in a helium atom from the 1s state to the 2s state. The same transition for the He+ ion requires approximately twice as much energy. Explain.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
- Q) A hydrogen atom emits radiation as a result of an electron transition to a lower energy level. Determine the highest frequency possible due to this transition if the atom emits a series of lines that lie in the visible part of the spectrum. Then, if the electron ends up in n = 1 level, prove that the atom emits a series of lines of wavelength that are not in the visible part of the spectrum.arrow_forwardIf, in 1 1 = Ry - you set ni = 1 and take n2 greater than 1, you generate what is known as the Lyman %3D series. Find the wavelength of the first mem- ber of this series. The value of ħ is 1.05457 × 10¬34 J.s; the Rydberg constant for hydrogen is 1.09735 × 10’ m¬'; the Bohr radius is 5.29177 × 10¬1" m; and the ground state energy for hydrogen is 13.6057 eV. Answer in units of nm. Consider the next three members of this se- ries. The wavelengths of successive members of the Lyman series approach a common limit as n2 → ∞. What is this limit? Answer in units of nm.arrow_forwardA photon is emitted during the transition from the n = 7 state to the n = 1 state in the hydrogen atom. Note: Rydberg constant: 2.18x10-18 J Can this photon eject an electron from cesium metal (work function of Cs is 3.42x10-19 J) and why? What would be the speed of that electron once is ejected (mass electron is 9.109x10-31 kg)?arrow_forward
- In a hydrogen atom, the small magnetic moment of the proton interacts with the magnetic moment of the electron. This results in a 5.87 μeV energy difference between the ms = + 1/2 and ms = -1/2 states. What wavelength photon is emitted in a transition between these two states?arrow_forwardElectron transitions for the Hydrogen atom n=7 n=6 n=5 n=4 n=3 Paschen series E(n) to E(n=3) n=2 Balmer series E(n) to E(n=2) n=1 Lyman series E(n) to E(n-1) The series limit wavelength of the Balmer series is emitted as the electron in the hydrogen atom falls from n = ∞ to n = 2. What would be the specific wavelength of such a line? [Hint :1/λ = RH(1/n/2-1/n2), RH being Rydberg constant = 1.097 x 107 /m] 560 nm 365 nm 400 nm 600 nm Brackett series E[n) to E(n=4)arrow_forwardIn a hydrogen atom, the small magnetic moment of the proton interacts with the magnetic moment of the electron. This results in a 5.87 meV energy difference between the ms = + 12 and ms = -12 states. What wavelength photon is emitted in a transition between these two states?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
University Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax