Fundamentals of Physics, Binder Ready Version
10th Edition
ISBN: 9781118230640
Author: Halliday, David; Resnick, Robert; Walker, Jearl
Publisher: WILEY
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Chapter 39, Problem 62P
To determine
To find:
a) The
b) The wavelength of the series limit.
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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) 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?
(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, Binder Ready Version
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
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- For the hydrogen atom, the Balmer Series is the series of line spectra produced when an election falls from any higher energy level and has n=2 as its final energy state. The longest wavelength is the Balmer Series is 656.7 nm. what was the initial principal energy level (n) associated with this photon. Will the energy value in the trasnistion be the geatest or smalles in the entie Balmer Series?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_forwardWhat is the wavelength of the hydrogen Balmer Series photon for m=4 and n=2 using the Rydberg forumla?arrow_forward
- A student in a physics laboratory observes a hydrogen spectrum with a diffraction grating for the purpose of measuring the wavelengths of the emitted radiation. In the spectrum, she observes a yellow line and finds its wavelength to be 589 nm. (a) Assuming this is part of the Balmer series, determine ni , the principal quantum number of the initial state. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?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_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
- 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 energy is required to remove the remaining…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 is, approximately, the longest wavelength that…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_forward
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