Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Question
Chapter 29, Problem 48P
To determine
The uncertainty in the radial position of the electron in ground state of the hydrogen atom.
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It may be argued on theoretical grounds that the radius of the hydrogen atom should depend only on the fundamental constants h, e, the electrostatic force constant k = 1/4πℰ0, and m (the electron’s mass). Use dimensional analysis to show that the combination of these factors that yields a result with dimensions of length is h2kme2.
The nucleus of a hydrogen atom is a single proton, which has a radius of about 1.1 × 10-15 m. The single electron in a
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to the density of the complete hydrogen atom?
Number i 1.12E+13
Units
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Chapter 39, Problem 043
In the ground state of the hydrogen atom, the electron has a total energy of -13.6 ev. What are (a) its kinetic energy and
(b) its potential energy if the electron is a distance 4.0a from the central nucleus? Here a is the Bohr radius.
(a) Number
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(b) Number
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Chapter 29 Solutions
Principles of Physics: A Calculus-Based Text
Ch. 29.2 - Prob. 29.1QQCh. 29.2 - Prob. 29.2QQCh. 29.4 - Prob. 29.3QQCh. 29.5 - Prob. 29.4QQCh. 29.6 - Prob. 29.5QQCh. 29.6 - Prob. 29.6QQCh. 29 - Prob. 1OQCh. 29 - Prob. 2OQCh. 29 - Prob. 3OQCh. 29 - Prob. 4OQ
Ch. 29 - Prob. 5OQCh. 29 - Prob. 6OQCh. 29 - Prob. 7OQCh. 29 - Prob. 8OQCh. 29 - Prob. 9OQCh. 29 - Prob. 10OQCh. 29 - Prob. 1CQCh. 29 - Prob. 2CQCh. 29 - Prob. 3CQCh. 29 - Prob. 4CQCh. 29 - Prob. 5CQCh. 29 - Prob. 6CQCh. 29 - Prob. 7CQCh. 29 - Prob. 8CQCh. 29 - Prob. 9CQCh. 29 - Prob. 10CQCh. 29 - Prob. 1PCh. 29 - Prob. 2PCh. 29 - Prob. 3PCh. 29 - Prob. 4PCh. 29 - Prob. 5PCh. 29 - Prob. 6PCh. 29 - Prob. 7PCh. 29 - Prob. 8PCh. 29 - Prob. 10PCh. 29 - Prob. 11PCh. 29 - Prob. 12PCh. 29 - Prob. 13PCh. 29 - Prob. 14PCh. 29 - Prob. 15PCh. 29 - Prob. 16PCh. 29 - Prob. 17PCh. 29 - Prob. 18PCh. 29 - Prob. 19PCh. 29 - Prob. 20PCh. 29 - Prob. 21PCh. 29 - Prob. 22PCh. 29 - Prob. 23PCh. 29 - Prob. 24PCh. 29 - Prob. 25PCh. 29 - Prob. 26PCh. 29 - Prob. 27PCh. 29 - Prob. 28PCh. 29 - Prob. 29PCh. 29 - Prob. 30PCh. 29 - Prob. 31PCh. 29 - Prob. 32PCh. 29 - Prob. 33PCh. 29 - Prob. 34PCh. 29 - Prob. 35PCh. 29 - Prob. 36PCh. 29 - Prob. 37PCh. 29 - Prob. 38PCh. 29 - Prob. 39PCh. 29 - Prob. 40PCh. 29 - Prob. 41PCh. 29 - Prob. 42PCh. 29 - Prob. 43PCh. 29 - Prob. 44PCh. 29 - Prob. 45PCh. 29 - Prob. 46PCh. 29 - Prob. 47PCh. 29 - Prob. 48PCh. 29 - Prob. 49PCh. 29 - Prob. 50PCh. 29 - Prob. 51PCh. 29 - Prob. 52PCh. 29 - Prob. 53PCh. 29 - Prob. 54PCh. 29 - Prob. 55PCh. 29 - Prob. 57PCh. 29 - Prob. 58PCh. 29 - Prob. 59PCh. 29 - Prob. 60PCh. 29 - Prob. 61PCh. 29 - Prob. 63PCh. 29 - Prob. 64PCh. 29 - Prob. 65PCh. 29 - Prob. 66P
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- Determine the distance between the electron and proton in an atom if the potential energy UU of the electron is 11 eV (electronvolt, 1 eV =1.6×10−19=1.6×10−19 J). Give your answer in Angstrom (1 A = 10-10 m).arrow_forwardA mercury atom emits light at many wavelengths, two of which are at (1) 404.7 nm and (2) 435.8 nm . Both of these transitions are to the same final state. Determine the energy of each emitted wavelength (in J). A E1 = 7.908 x 10-19 J and E2 = 6.558 x10-19 J E1 = 4.628 x 10-19 J and E2 = 4.968 x10-19 J E1 = 5.908 x 10-19 J and E2 = 5.558 x10-19J E1 = 4.908 x 10-19 J and E2 = 4.558 x10-19 Jarrow_forwardUse the table to determine the energy in eV of the photon emitted when an electron jumps down from the n = 2 orbit to the n = 1 orbit of a hydrogen atom. Allowed Values of the Hydrogen Electron's Radius and Energy for Low Values of n n rn En 1 0.053 nm −13.60 eV 2 0.212 nm −3.40 eV 3 0.477 nm −1.51 eV 4 0.848 nm −0.85 eV eVarrow_forward
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