Modern Physics
3rd Edition
ISBN: 9781111794378
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
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Question
Chapter 6, Problem 3P
(a)
To determine
The de Broglie wavelength of electron.
(b)
To determine
The momentum of electron.
(c)
To determine
The energy of electron in electron volts.
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A free electron has a wave function ψ(x) = Ae i(5.00×1010x)where x is in meters. Find its (a) de Broglie wavelength, (b) momentum, and (c) kinetic energy in electron volts.
An atom in an excited state of 4.7 eV emits a photon and ends up in the ground state. The lifetime of the excited state is 1.0 x 10-13 s. (a) What is the energy uncertainty of the emitted photon? (b) What is the spectral line width (in wavelength) of the photon?
A free electron moving along the x-direction (one for which V(x) = 0) would have a wave
function of the form
f(x) = A eikx + B e−ikx
where A and B are constants.
(a) If the wavelength of this wave function (in radians) equals the de Broglie wavelength of
the electron, and its velocity, v = 8.40 × 10² m/sec, what is the value of k (in nm¯¹)?
Express your answer in scientific notation with three significant figures.
(b) The Hamiltonian operator for a free electron is given by
p²
ħ² d²
Ĥ
2me
2me dx²
The wave function provided at the top is an eigenstate of Â. If one measures the energy
for an electron in this state using Â, what would be the result, and how does it compare
to the classical kinetic energy of a free electron with this velocity?
Chapter 6 Solutions
Modern Physics
Ch. 6.4 - Prob. 1ECh. 6.4 - Prob. 2ECh. 6.5 - Prob. 4ECh. 6.7 - Prob. 5ECh. 6.8 - Prob. 6ECh. 6 - Prob. 1QCh. 6 - Prob. 2QCh. 6 - Prob. 3QCh. 6 - Prob. 4QCh. 6 - Prob. 5Q
Ch. 6 - Prob. 6QCh. 6 - Prob. 7QCh. 6 - Prob. 8QCh. 6 - Prob. 1PCh. 6 - Prob. 2PCh. 6 - Prob. 3PCh. 6 - Prob. 5PCh. 6 - Prob. 6PCh. 6 - Prob. 7PCh. 6 - Prob. 8PCh. 6 - Prob. 9PCh. 6 - Prob. 10PCh. 6 - Prob. 11PCh. 6 - Prob. 12PCh. 6 - Prob. 13PCh. 6 - Prob. 14PCh. 6 - Prob. 15PCh. 6 - Prob. 16PCh. 6 - Prob. 17PCh. 6 - Prob. 18PCh. 6 - Prob. 19PCh. 6 - Prob. 21PCh. 6 - Prob. 24PCh. 6 - Prob. 25PCh. 6 - Prob. 26PCh. 6 - Prob. 28PCh. 6 - Prob. 29PCh. 6 - Prob. 30PCh. 6 - Prob. 31PCh. 6 - Prob. 32PCh. 6 - Prob. 33PCh. 6 - Prob. 34PCh. 6 - Prob. 35PCh. 6 - Prob. 37PCh. 6 - Prob. 38P
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- A free electron moving along the x-direction (one for which V(x) = 0) would have a wave function of the form f(x) = A eikx + Be -ikx where A and B are constants. (a) If the wavelength of this wave function (in radians) equals the de Broglie wavelength of the electron, and its velocity, v = 8.40 x 10² m/sec, what is the value of k (in nm-¹)? Express your answer in scientific notation with three significant figures. (b) The Hamiltonian operator for a free electron is given by ħ² d² p² 2me 2m₂ dx² The wave function provided at the top is an eigenstate of Â. If one measures the energy for an electron in this state using Â, what would be the result, and how does it compare to the classical kinetic energy of a free electron with this velocity? Ĥ = =arrow_forwardAn electron has a momentum py = 1.40×10−251.40×10−25 kg.m/s. What is the minimum uncertainty in its position that will keep the relative uncertainty in its momentum (Δpy/p) below 2.7%?arrow_forwardDuring a certain experiment, the de Broglie wavelength of an electron is 400 nm = 4.0 ✕ 10−7 m, which is the same as the wavelength of violet light. How fast (in m/s) is the electron moving?arrow_forward
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- The electron beam in a scanning electron microscope (SEM) can have an energy ranging from 0.2 keV to 40 keV. (Note: an eV is an electronvolt, the measure of an amount of kinetic energy for a single electron. 1 eV = 1.60218×10−19 J and 1 J = 1 kg m2 s−2). What is the de Broglie wavelength of the electron beam in the SEM when the electrons are accelerated to kinetic energy of 6.6 keV?arrow_forwardThe speed of an electron is measured to within an uncertainty of 2.0 × 104 m/s. What is the size of the smallest region of space in which the electron can be confined?arrow_forwardA free electron has a wave function ψ(x)= Asin (5x1010 x) where x is measured in meters. Find the electron's de Broglie wavelength the electron's momentumarrow_forward
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