EBK PHYSICS
5th Edition
ISBN: 8220103026918
Author: Walker
Publisher: PEARSON
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Question
Chapter 31, Problem 28PCE
(a)
To determine
The shortest wavelength of the Lyman series for
Be 3 +
.
(b)
To determine
The ionization energy required to remove the final electron in
Be 3 +
.
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If, 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.
The 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 region
What is the wavelength of the hydrogen Balmer Series photon for m=4 and n=2 using the Rydberg forumla?
Chapter 31 Solutions
EBK PHYSICS
Ch. 31.1 - Prob. 1EYUCh. 31.2 - Prob. 2EYUCh. 31.3 - Prob. 3EYUCh. 31.4 - Prob. 4EYUCh. 31.5 - Prob. 5EYUCh. 31.6 - Prob. 6EYUCh. 31.7 - Prob. 7EYUCh. 31 - Prob. 1CQCh. 31 - Prob. 2CQCh. 31 - Prob. 3CQ
Ch. 31 - Prob. 4CQCh. 31 - Prob. 5CQCh. 31 - Prob. 6CQCh. 31 - Prob. 7CQCh. 31 - Prob. 8CQCh. 31 - Prob. 9CQCh. 31 - Prob. 1PCECh. 31 - Prob. 2PCECh. 31 - Prob. 3PCECh. 31 - Prob. 4PCECh. 31 - Prob. 5PCECh. 31 - Prob. 6PCECh. 31 - Prob. 7PCECh. 31 - Prob. 8PCECh. 31 - Prob. 9PCECh. 31 - Prob. 10PCECh. 31 - Prob. 11PCECh. 31 - Prob. 12PCECh. 31 - Prob. 13PCECh. 31 - Prob. 14PCECh. 31 - Prob. 15PCECh. 31 - Prob. 16PCECh. 31 - Prob. 17PCECh. 31 - Prob. 18PCECh. 31 - Prob. 19PCECh. 31 - Prob. 20PCECh. 31 - Prob. 21PCECh. 31 - Prob. 22PCECh. 31 - Prob. 23PCECh. 31 - Prob. 24PCECh. 31 - Prob. 25PCECh. 31 - Prob. 26PCECh. 31 - Prob. 27PCECh. 31 - Prob. 28PCECh. 31 - Prob. 29PCECh. 31 - Prob. 30PCECh. 31 - Prob. 31PCECh. 31 - Prob. 32PCECh. 31 - Prob. 33PCECh. 31 - Prob. 34PCECh. 31 - Prob. 35PCECh. 31 - Prob. 36PCECh. 31 - Prob. 37PCECh. 31 - Prob. 38PCECh. 31 - Prob. 39PCECh. 31 - Prob. 40PCECh. 31 - Prob. 41PCECh. 31 - Prob. 42PCECh. 31 - Prob. 43PCECh. 31 - Prob. 44PCECh. 31 - Prob. 45PCECh. 31 - Prob. 46PCECh. 31 - Prob. 47PCECh. 31 - Prob. 48PCECh. 31 - Prob. 49PCECh. 31 - Prob. 50PCECh. 31 - Prob. 51PCECh. 31 - Prob. 52PCECh. 31 - Give the electronic configuration for the ground...Ch. 31 - Prob. 54PCECh. 31 - Prob. 55PCECh. 31 - Prob. 56PCECh. 31 - The configuration of the outer electrons in Ni is...Ch. 31 - Prob. 58PCECh. 31 - Prob. 59PCECh. 31 - Prob. 60PCECh. 31 - Prob. 61PCECh. 31 - Prob. 62PCECh. 31 - Prob. 63PCECh. 31 - Prob. 64PCECh. 31 - Prob. 65PCECh. 31 - Prob. 66PCECh. 31 - Prob. 67PCECh. 31 - Prob. 68GPCh. 31 - Prob. 69GPCh. 31 - Prob. 70GPCh. 31 - Prob. 71GPCh. 31 - Prob. 72GPCh. 31 - Prob. 73GPCh. 31 - Prob. 74GPCh. 31 - Prob. 75GPCh. 31 - Prob. 76GPCh. 31 - Prob. 77GPCh. 31 - Prob. 78GPCh. 31 - Prob. 79GPCh. 31 - Prob. 80GPCh. 31 - Prob. 81GPCh. 31 - Prob. 82GPCh. 31 - Prob. 83GPCh. 31 - Prob. 84PPCh. 31 - Prob. 85PPCh. 31 - Prob. 86PPCh. 31 - Prob. 87PPCh. 31 - Prob. 88PPCh. 31 - Prob. 89PP
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- If an atom has an election in the n = 5 state with m = 3, what are the possible values of l?arrow_forwardDerive an expression for the ratio of X-ray photon frequency for two elements with atomic numbers Z1 and Z2.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_forward
- The work function for potassium is 2.26 eV. What is the cutoff frequency when this metal is used as photoelectrode? What is the stopping potential when for the emitted electrons when this photo electrode is exposed to radiation of frequency 1200 THz?arrow_forwarda) An electron in a hydrogen atom has energy E= -3.40 eV, where the zero of energy is at the ionization threshold. In the Bohr model, what is the angular momentum of the electron? Express your result as a multiple of ħ. Ans. b) What is the deBroglie wavelength of the electron when it is in this state? Ans. c) When the electron is in this state, what is the ratio of the circumference of the orbit of the electron to the deBroglie wavelength of the electron? Ans. d) The electron makes a transition from the state with energy E= -3.40 eV to the ground state, that has energy -13.6 eV. What is the wavelength of the photon emitted during this transition? Ans.arrow_forwardShow that the longest wavelength of the Balmer series and the longest two wavelengths of the Lyman series sat- isfy the Ritz combination principle. For the Lyman series, limit = 91.13 nm.arrow_forward
- 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 eVarrow_forwardThe x-ray spectrum is for 35.0 keV electrons striking a molybdenum (Z= 42) target. If you substitute a silver (Z = 47) target for the molybdenum target, will (a) lmin, (b) the wavelength for the Ka line, and (c) the wavelength for the Kb line increase, decrease, or remain unchanged?arrow_forwardThe visible lines of the Balmer series were observed first because they are most easily seen. Show that the wavelengths of spectral lines in the Lyman (n = 1) and Paschen (n = 3) series are not in the visible region. Find the wavelengths of the four visible atomic hydrogen lines. Assume the visible wavelength region is λ = 400– 700 nm.arrow_forward
- (a) A hydrogen atom has its electron in the n = 2 level. The radius of the electron's orbit in the Bohr model is 0.212 nm. Find the de Broglie wavelength of the electron under these circumstances. (b) What is the momentum, mv, of the electron in its orbit? kg-m/sarrow_forwardAssuming that only a single electron is present and a Bohr model, calculate the mean radius, orbital velocity, and energy of a N=1 electron of hydrogen N=4 electron for lead N=1 electron for plutonium N=1 electron for an element with Z = 142arrow_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_forward
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