Fundamentals of Physics Extended
10th Edition
ISBN: 9781118230725
Author: David Halliday, Robert Resnick, Jearl Walker
Publisher: Wiley, John & Sons, Incorporated
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 39, Problem 62P
To determine
To find:
a) The
b) The wavelength of the series limit.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
A) What is the least amount of energy, in electron volts, that must be given to a hydrogen atom which is initially in its ground level so that it can emit the HαHα line in the Balmer series?
Express your answer in electronvolts to three significant figures.
B) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the n = 3 level and eventually ends up in the ground level?
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
(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.
Chapter 39 Solutions
Fundamentals of Physics Extended
Ch. 39 - Prob. 1QCh. 39 - Prob. 2QCh. 39 - Prob. 3QCh. 39 - Prob. 4QCh. 39 - Prob. 5QCh. 39 - Prob. 6QCh. 39 - Prob. 7QCh. 39 - Prob. 8QCh. 39 - Prob. 9QCh. 39 - Prob. 10Q
Ch. 39 - Prob. 11QCh. 39 - Prob. 12QCh. 39 - Prob. 13QCh. 39 - Prob. 14QCh. 39 - Prob. 15QCh. 39 - Prob. 1PCh. 39 - Prob. 2PCh. 39 - Prob. 3PCh. 39 - Prob. 4PCh. 39 - Prob. 5PCh. 39 - Prob. 6PCh. 39 - Prob. 7PCh. 39 - Prob. 8PCh. 39 - Prob. 9PCh. 39 - Prob. 10PCh. 39 - Prob. 11PCh. 39 - Prob. 12PCh. 39 - Prob. 13PCh. 39 - Prob. 14PCh. 39 - Prob. 15PCh. 39 - Prob. 16PCh. 39 - Prob. 17PCh. 39 - Prob. 18PCh. 39 - Prob. 19PCh. 39 - Prob. 20PCh. 39 - Prob. 21PCh. 39 - Prob. 22PCh. 39 - Prob. 23PCh. 39 - Prob. 24PCh. 39 - Prob. 25PCh. 39 - Prob. 26PCh. 39 - Prob. 27PCh. 39 - Prob. 28PCh. 39 - Prob. 29PCh. 39 - Prob. 30PCh. 39 - Prob. 31PCh. 39 - Prob. 32PCh. 39 - Prob. 33PCh. 39 - Prob. 34PCh. 39 - Prob. 35PCh. 39 - Prob. 36PCh. 39 - Prob. 37PCh. 39 - Prob. 38PCh. 39 - Prob. 39PCh. 39 - Prob. 40PCh. 39 - Prob. 41PCh. 39 - Prob. 42PCh. 39 - Prob. 43PCh. 39 - Prob. 44PCh. 39 - Prob. 45PCh. 39 - Prob. 46PCh. 39 - Prob. 47PCh. 39 - Prob. 48PCh. 39 - Prob. 49PCh. 39 - Prob. 50PCh. 39 - Prob. 51PCh. 39 - Prob. 52PCh. 39 - Prob. 53PCh. 39 - Prob. 54PCh. 39 - Prob. 55PCh. 39 - Prob. 56PCh. 39 - Prob. 57PCh. 39 - Prob. 58PCh. 39 - Prob. 59PCh. 39 - Prob. 60PCh. 39 - Prob. 61PCh. 39 - Prob. 62PCh. 39 - Prob. 63PCh. 39 - Prob. 64PCh. 39 - A diatomic gas molcculc consistsof two atoms of...Ch. 39 - Prob. 66PCh. 39 - Prob. 67PCh. 39 - Prob. 68PCh. 39 - Prob. 69PCh. 39 - Prob. 70PCh. 39 - An old model of a hydrogen atom has the charge e...Ch. 39 - Prob. 72PCh. 39 - Prob. 73P
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
- 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_forwardWhat is the wavelength of the hydrogen Balmer Series photon for m=4 and n=2 using the Rydberg forumla?arrow_forwardIn a normal Zeeman Effect experiment, spectral splitting of the line at the wavelength 643.8 nm corresponding to the transition 5'D, → 5'P, of cadmium atoms is to be observed. The spectrometer has a resolution of 0.01 nm. Minimum magnetic field needed to observe this is (m. = 9.1x10-' kg,e =1.6x-19 C,c = 3×10° m/s) -31 (а) 0.26T (b) 0.527 (c) 2.6T (d) 5.27arrow_forward
- A Hydrogen atom initially in its ground state i.e., n = 1 level, absorbs a photon and ends up in n = 4 level. (a) What must have been the frequency of the photon? Now the electron makes spontaneous emission and comes back to the ground state. (b) What are the possible frequencies of the photons emitted during this process?arrow_forwardSingly ionized helium has a single orbiting electron, so the mathematicsof the Bohr hydrogen atom will apply, with one important difference: The charge of the nucleus is twice that of the single proton at the center of a hydrogen atom. This changes the energy levels; the magnitude of each energy is greater than the corresponding Bohr level by a factor of 22 = 4: The Balmer and Lyman series of spectral lines in hydrogen have analogs in singly ionized helium, but at shorter wavelengths; the photons corresponding to these transitions are beyond the visiblelight spectrum. The transitions that end on the n = 4 state produce a set of spectral lines called the Pickering series. The visible-light lines in this series were first seen in the light from certain hot stars, but some of the lines overlap the hydrogen Balmer series lines, so these lines were initially missed. This led to an initial mischaracterization of the source of the lines. The longest wavelength in the hydrogen Balmer series…arrow_forwardSingly ionized helium has a single orbiting electron, so the mathematicsof the Bohr hydrogen atom will apply, with one important difference: The charge of the nucleus is twice that of the single proton at the center of a hydrogen atom. This changes the energy levels; the magnitude of each energy is greater than the corresponding Bohr level by a factor of 22 = 4: The Balmer and Lyman series of spectral lines in hydrogen have analogs in singly ionized helium, but at shorter wavelengths; the photons corresponding to these transitions are beyond the visiblelight spectrum. The transitions that end on the n = 4 state produce a set of spectral lines called the Pickering series. The visible-light lines in this series were first seen in the light from certain hot stars, but some of the lines overlap the hydrogen Balmer series lines, so these lines were initially missed. This led to an initial mischaracterization of the source of the lines. The Paschen series of wavelengths in the hydrogen…arrow_forward
- (a) If one subshell of an atom has 9 electrons in it, what is the minimum value of l ? (b) What is the spectroscopic notation for this atom, if this subshell is part of the n = 3shell?arrow_forwarda) Consider that the first line of the Balmer series of the 1H spectrum is due to the transitions between the states described by the 3p and 2s orbitals. This line suffers unfolding when the system is subject to an intense magnetic fieldexternal. Calculate the wave numbers (in cm-1 ) of the lines observed when a10,0000T magnetic field is applied. (hint: pay attention to the numberssignificant and approximations made in the calculation steps)arrow_forwardSingly ionized helium has a single orbiting electron, so the mathematicsof the Bohr hydrogen atom will apply, with one important difference: The charge of the nucleus is twice that of the single proton at the center of a hydrogen atom. This changes the energy levels; the magnitude of each energy is greater than the corresponding Bohr level by a factor of 22 = 4: The Balmer and Lyman series of spectral lines in hydrogen have analogs in singly ionized helium, but at shorter wavelengths; the photons corresponding to these transitions are beyond the visiblelight spectrum. The transitions that end on the n = 4 state produce a set of spectral lines called the Pickering series. The visible-light lines in this series were first seen in the light from certain hot stars, but some of the lines overlap the hydrogen Balmer series lines, so these lines were initially missed. This led to an initial mischaracterization of the source of the lines. What energy is required to remove the remaining…arrow_forward
- Singly ionized helium has a single orbiting electron, so the mathematicsof the Bohr hydrogen atom will apply, with one important difference: The charge of the nucleus is twice that of the single proton at the center of a hydrogen atom. This changes the energy levels; the magnitude of each energy is greater than the corresponding Bohr level by a factor of 22 = 4: The Balmer and Lyman series of spectral lines in hydrogen have analogs in singly ionized helium, but at shorter wavelengths; the photons corresponding to these transitions are beyond the visiblelight spectrum. The transitions that end on the n = 4 state produce a set of spectral lines called the Pickering series. The visible-light lines in this series were first seen in the light from certain hot stars, but some of the lines overlap the hydrogen Balmer series lines, so these lines were initially missed. This led to an initial mischaracterization of the source of the lines. What is, approximately, the longest wavelength that…arrow_forward(a) Using de-Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom. (b) The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the state?arrow_forward(4) Electronic energy level of a hydrogen atom is given by R ; п %3D 1,2, 3,... n2 E = - and R = 13.6 eV. Each energy level has degeneracy 2n2 (degeneracy is the number of equivalent configurations associated with the energy level). (a) Derive the partition function for a hydrogen atom at a constant temperature. (b) Consider that the energy level of a hydrogen atom is approximated by a two level system, n = 1,2. Estimate the mean energy at 300 K.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning