Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
9th Edition
ISBN: 9781305266292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Question
Chapter 42, Problem 20P
(a)
To determine
Prove that the period of electron is
(b)
To determine
Number of revolutions made by electron in the excited state.
(c)
To determine
Check whether is it possible to treat the electron in
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An electron is in the nth Bohr orbit of the hydrogen atom. (a) Show that the period of the electron is T = n3t0 and determine the numerical value of t0. (b) On average, an electron remains in the n = 2 orbit for approximately 10 ms before it jumps down to the n = 1 (ground-state) orbit. How many revolutions does the electron make in the excited state? (c) Define the period of one revolution as an electron year, analogous to an Earth year being the period of the Earth’s motion around the Sun. Explain whether we should think of the electron in the n = 2 orbit as “living for a long time.”
An electron is in the nth Bohr orbit of the hydrogen atom.
n3
(a) Show that the period of the electron is T = to n³ and determine the numerical value of to.
153
as
(b) On average, an electron remains in the n = 2 orbit for approximately 8 us before it jumps down to the n = 1
(ground-state)orbit. How many revolutions does the electron make in the excited state?
8.26e+09
×
(c) Define the period of one revolution as an electron year, analogous to an Earth year being the period of the
Earth's motion around the Sun. Explain whether we should think of the electron in the n = 2 orbit as "living for a
long time."
What is the average radius of the orbit of an electron in the n=2 energy level of an oxygen atom (Z=8)? Express your answer in pico-meters.
Chapter 42 Solutions
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
Ch. 42.3 - Prob. 42.1QQCh. 42.3 - Prob. 42.2QQCh. 42.4 - Prob. 42.3QQCh. 42.4 - Prob. 42.4QQCh. 42.8 - Prob. 42.5QQCh. 42 - Prob. 1OQCh. 42 - Prob. 2OQCh. 42 - Prob. 3OQCh. 42 - Prob. 4OQCh. 42 - Prob. 5OQ
Ch. 42 - Prob. 6OQCh. 42 - Prob. 7OQCh. 42 - Prob. 8OQCh. 42 - Prob. 9OQCh. 42 - Prob. 10OQCh. 42 - Prob. 11OQCh. 42 - Prob. 12OQCh. 42 - Prob. 13OQCh. 42 - Prob. 14OQCh. 42 - Prob. 15OQCh. 42 - Prob. 1CQCh. 42 - Prob. 2CQCh. 42 - Prob. 3CQCh. 42 - Prob. 4CQCh. 42 - Prob. 5CQCh. 42 - Prob. 6CQCh. 42 - Prob. 7CQCh. 42 - Prob. 8CQCh. 42 - Prob. 9CQCh. 42 - Prob. 10CQCh. 42 - Prob. 11CQCh. 42 - Prob. 12CQCh. 42 - Prob. 1PCh. 42 - Prob. 2PCh. 42 - Prob. 3PCh. 42 - Prob. 4PCh. 42 - Prob. 5PCh. 42 - Prob. 6PCh. 42 - Prob. 7PCh. 42 - Prob. 8PCh. 42 - Prob. 9PCh. 42 - Prob. 10PCh. 42 - Prob. 11PCh. 42 - Prob. 12PCh. 42 - Prob. 13PCh. 42 - Prob. 14PCh. 42 - Prob. 15PCh. 42 - Prob. 16PCh. 42 - Prob. 17PCh. 42 - Prob. 18PCh. 42 - Prob. 19PCh. 42 - Prob. 20PCh. 42 - Prob. 21PCh. 42 - Prob. 23PCh. 42 - Prob. 24PCh. 42 - Prob. 25PCh. 42 - Prob. 26PCh. 42 - Prob. 27PCh. 42 - Prob. 28PCh. 42 - Prob. 29PCh. 42 - Prob. 30PCh. 42 - Prob. 31PCh. 42 - Prob. 32PCh. 42 - Prob. 33PCh. 42 - Prob. 34PCh. 42 - Prob. 35PCh. 42 - Prob. 36PCh. 42 - Prob. 37PCh. 42 - Prob. 38PCh. 42 - Prob. 39PCh. 42 - Prob. 40PCh. 42 - Prob. 41PCh. 42 - Prob. 43PCh. 42 - Prob. 44PCh. 42 - Prob. 45PCh. 42 - Prob. 46PCh. 42 - Prob. 47PCh. 42 - Prob. 48PCh. 42 - Prob. 49PCh. 42 - Prob. 50PCh. 42 - Prob. 51PCh. 42 - Prob. 52PCh. 42 - Prob. 53PCh. 42 - Prob. 54PCh. 42 - Prob. 55PCh. 42 - Prob. 56PCh. 42 - Prob. 57PCh. 42 - Prob. 58PCh. 42 - Prob. 59PCh. 42 - Prob. 60PCh. 42 - Prob. 61PCh. 42 - Prob. 62PCh. 42 - Prob. 63PCh. 42 - Prob. 64PCh. 42 - Prob. 65APCh. 42 - Prob. 66APCh. 42 - Prob. 67APCh. 42 - Prob. 68APCh. 42 - Prob. 69APCh. 42 - Prob. 70APCh. 42 - Prob. 71APCh. 42 - Prob. 72APCh. 42 - Prob. 73APCh. 42 - Prob. 74APCh. 42 - Prob. 75APCh. 42 - Prob. 76APCh. 42 - Prob. 77APCh. 42 - Prob. 78APCh. 42 - Prob. 79APCh. 42 - Prob. 80APCh. 42 - Prob. 81APCh. 42 - Prob. 82APCh. 42 - Prob. 83APCh. 42 - Prob. 84APCh. 42 - Prob. 85APCh. 42 - Prob. 86APCh. 42 - Prob. 87APCh. 42 - Prob. 88APCh. 42 - Prob. 89CPCh. 42 - Prob. 90CPCh. 42 - Prob. 91CP
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- For an electron in a hydrogen atom in the n=2 state, compute: (a) the angular momentum; (b) the kinetic energy; (c) the potential energy; and (d) the total energy.arrow_forwardThe figure shows a model of the energy levels of an atom. The atom is initially in state W, which is the ground state for the atom. After a short amount of time, the atom then transitions to state X. The atom then transitions to state Y before transitioning to state Z. The atom then transitions back to state W. Which of the following descriptions is correct about the atom as it transitions from state W to each subsequent state until it finally returns to its original state?arrow_forward(a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels. (c) The average lifetime of the first excited level of a hydrogen atom is 1.0 * 10-8 s. In the Bohr model, how many orbits does an electron in the n = 2 level complete before returning to the ground level?arrow_forward
- Determine the distance between the electron and proton in an atom if the potential energy U of the electron is 10.1 eV (electronvolt, 1 eV = 1.6 × 10-19 J). Give your answer in Angstrom (1 A = 10-10 m). Answer: Choose... +arrow_forwardConsidering the Bohr’s model, given that an electron is initially located at the ground state (n=1n=1) and it absorbs energy to jump to a particular energy level (n=nxn=nx). If the difference of the radius between the new energy level and the ground state is rnx−r1=5.247×10−9rnx−r1=5.247×10−9, determine nxnx and calculate how much energy is absorbed by the electron to jump to n=nxn=nx from n=1n=1. A. nx=9nx=9; absorbed energy is 13.4321 eV B. nx=10nx=10; absorbed energy is 13.464 eV C. nx=8nx=8; absorbed energy is 13.3875 eV D. nx=20nx=20; absorbed energy is 13.566 eV E. nx=6nx=6; absorbed energy is 13.22 eV F. nx=2nx=2; absorbed energy is 10.2 eV G. nx=12nx=12; absorbed energy is 13.506 eV H. nx=7nx=7; absorbed energy is 13.322 eVarrow_forwardIn the mid-1800s, scientists observed four visible lines in the emission spectrum of the hydrogen atom. These lines had wavelengths of 410 nm, 434 nm, 486 nm, and 656 nm, where 1 nm = 10-⁹ m. This series of lines is part of what became known as the Balmer series, after Johann Balmer demonstrated that their wavelengths A could be described by the equation 1 2 = 1.1 x 107 Although this equation worked, it was some time before scientists understood why it worked. In the early 1900s, Niels Bohr proposed that the electron in a hydrogen atom could exist only in certain quantized energy states. These states were given by the equation E = -13.6/n² eV. He further proposed that when an electron moved from a higher energy state to a lower one, the excess energy was released as a packet, or quantum, of light known as a photon. The corresponding wavelength of a photon could then be found with the equation λ = hc/E, where E is the energy of the photon, h = 4.1 x 10-15 eV's is Planck's constant, and…arrow_forward
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