Fundamentals of Physics, Volume 1, Chapter 1-20
10th Edition
ISBN: 9781118233764
Author: David Halliday
Publisher: WILEY
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 38, Problem 83P
To determine
To find
a) Average kinetic energy of neutron
b) de Broglie wavelength
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Neutrons in thermal equilibrium with matter have an average kinetic energy of (3/2)kT, where k is the Boltzmann constant and T, which may be taken to be 300 K, is the temperature of the environment of the neutrons. (a) What is the average kinetic energy of such a neutron? (b) What is the corresponding de Broglie wavelength?
The neutron has a mass of 1.67 ✕ 10-27 kg. Neutrons emitted in nuclear reactions can be slowed down via collisions with matter. They are referred to as thermal neutrons once they come into thermal equilibrium with their surroundings. The average kinetic energy (3kB T/2) of a thermal neutron is approximately 0.04 eV.Calculate the de Broglie wavelength of a neutron with a kinetic energy of 0.0980 eV.
The neutron has a mass of 1.67 × 10-27 kg. Neutrons emitted in nuclear reactions can be slowed down by collisions with matter. They are referred to as thermal neutrons after they come into thermal equilibrium with the environment. The average kinetic energy (3/2 kBT) of a thermal neutron isapproximately 0.04 eV. (a) Calculate the de Broglie wavelength of a neutron with a kinetic energy of 0.040 0 eV. (b) How does your answer compare with the characteristic atomic spacing in a crystal? (c) Explain whether you expect thermal neutrons to exhibit diffraction effects when scattered by a crystal.
Chapter 38 Solutions
Fundamentals of Physics, Volume 1, Chapter 1-20
Ch. 38 - Prob. 1QCh. 38 - Prob. 2QCh. 38 - Prob. 3QCh. 38 - Prob. 4QCh. 38 - Prob. 5QCh. 38 - Prob. 6QCh. 38 - Prob. 7QCh. 38 - Prob. 8QCh. 38 - Prob. 9QCh. 38 - Prob. 10Q
Ch. 38 - Prob. 11QCh. 38 - Prob. 12QCh. 38 - Prob. 13QCh. 38 - Prob. 14QCh. 38 - Prob. 15QCh. 38 - Prob. 16QCh. 38 - Prob. 1PCh. 38 - Prob. 2PCh. 38 - Prob. 3PCh. 38 - Prob. 4PCh. 38 - Prob. 5PCh. 38 - Prob. 6PCh. 38 - Prob. 7PCh. 38 - Prob. 8PCh. 38 - Prob. 9PCh. 38 - Prob. 10PCh. 38 - Prob. 11PCh. 38 - Prob. 12PCh. 38 - Prob. 13PCh. 38 - Prob. 14PCh. 38 - Prob. 15PCh. 38 - Prob. 16PCh. 38 - Prob. 17PCh. 38 - Prob. 18PCh. 38 - Prob. 19PCh. 38 - Prob. 20PCh. 38 - Prob. 21PCh. 38 - Prob. 22PCh. 38 - Prob. 23PCh. 38 - Prob. 24PCh. 38 - Prob. 25PCh. 38 - Prob. 26PCh. 38 - Prob. 27PCh. 38 - Prob. 28PCh. 38 - Prob. 29PCh. 38 - Prob. 30PCh. 38 - Prob. 31PCh. 38 - Prob. 32PCh. 38 - Prob. 33PCh. 38 - Prob. 34PCh. 38 - Prob. 35PCh. 38 - Prob. 36PCh. 38 - Prob. 37PCh. 38 - Prob. 38PCh. 38 - Prob. 39PCh. 38 - Prob. 40PCh. 38 - Prob. 41PCh. 38 - Prob. 42PCh. 38 - Prob. 43PCh. 38 - Prob. 44PCh. 38 - Prob. 45PCh. 38 - Prob. 46PCh. 38 - Prob. 47PCh. 38 - Prob. 48PCh. 38 - Prob. 49PCh. 38 - Prob. 50PCh. 38 - Prob. 51PCh. 38 - Prob. 52PCh. 38 - Prob. 53PCh. 38 - Prob. 54PCh. 38 - Prob. 55PCh. 38 - Prob. 56PCh. 38 - Prob. 57PCh. 38 - Prob. 58PCh. 38 - Prob. 59PCh. 38 - Prob. 60PCh. 38 - Prob. 61PCh. 38 - Prob. 62PCh. 38 - Prob. 63PCh. 38 - Prob. 64PCh. 38 - Prob. 65PCh. 38 - Prob. 66PCh. 38 - Prob. 67PCh. 38 - Prob. 68PCh. 38 - Prob. 69PCh. 38 - Prob. 70PCh. 38 - Prob. 71PCh. 38 - Prob. 72PCh. 38 - Prob. 73PCh. 38 - Prob. 74PCh. 38 - Prob. 75PCh. 38 - Prob. 76PCh. 38 - Prob. 77PCh. 38 - Prob. 78PCh. 38 - Prob. 79PCh. 38 - Prob. 80PCh. 38 - Prob. 81PCh. 38 - Prob. 82PCh. 38 - Prob. 83PCh. 38 - Prob. 84PCh. 38 - Prob. 85PCh. 38 - Prob. 86PCh. 38 - Prob. 87PCh. 38 - Prob. 88PCh. 38 - Prob. 89PCh. 38 - Prob. 90P
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 de Broglie wavelength of an electron has to do with spatial resolution of an electron microscope, which is often expressed in the unit of length Å (Angstrom). The 1 V potential difference causes an electron to gain kinetic energy EK of 1 electron Volt (eV). In the SI units, kinetic energy in eV must be converted to Joules. The conversion factor is 1 eV = 1.6 x 10-19 Joule. And, the formula for the wavelength is: λ = h / p = h / √(2 m EK) where m is electron mass. Calculate the de Broglie wavelength of an electron (in Å) when the electron is accelerated from rest through a potential difference of: a) 1 kV = 1,000 V (a low resolution setting of microscope), b) 10 kV = 10,000 V (intermediate resolution). c) 100 kV = 100,000 V (high resolution),arrow_forwardThe average energy of a photon in a pulsed laser beam is 2.39 eV, with a minimum uncertainty of 0.0155 eV. Each pulse has an average of 5.00 x 1012 photons. Find (a) the time duration of each pulse, (b) the wavelength of the light, and (c) the energy per pulse in J.arrow_forwardA photon with wavelength X scatters off an electron at rest, at an angle with the incident direction. The Compton wavelength of the electron Ac = 0.0024 nm. a) For λ = 0.0006 nm and 0 = 53 degrees, find the wavelength X' of the scattered photon in nanometres. b) Obtain a formula for the energy of the electron Ee after collision, in terms of the universal constants h, c and the variables X, X' and Ac. The answer must be expressed in terms of these variables only. (Please enter an algebraic expression using latex format; do not input any numerical values) c) Using the energy conservation condition, find the value of the electron energy Ee after scattering in units of keV. d) Write an algebraic expression for the electron's momentum pe in terms of its energy Ee, its mass me and the speed of light c. e) What is the de Broglie wavelength of the scattered electron ? Express your answer in terms of Ee, me, and X and c. f) Find the value of the de Broglie wavelength of the scattered electron…arrow_forward
- The radii of atomic nuclei are of the order of 5.0 * 10-15 m. (a) Estimate the minimum uncertainty in the momentum of a proton if it is confined within a nucleus. (b) Take this uncertainty in momentum to be an estimate of the mag- nitude of the momentum. Use the relativistic relationship between energy and momentum, Eq. (37.39), to obtain an estimate of the ki- netic energy of a proton confined within a nucleus. (c) For a proton to remain bound within a nucleus, what must the magnitude of the (negative) potential energy for a proton be within the nucleus? Give your answer in eV and in MeV. Compare to the potential energy for an electron in a hydrogen atom, which has a magnitude of a few tens of eV. (This shows why the interaction that binds the nucleus together is called the “strong nuclear force.”)arrow_forwardThe 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 existence of the atomic nucleus was discovered in 1911 by Ernest Rutherford, who properly interpreted some experiments in which a beam of alpha particles was scattered from a metal foil of atoms such as gold. (a) If the alpha particles had a kinetic energy of 7.5 MeV, what was their de Broglie wavelength? (b) Explain whether the wave nature of the incident alpha particles should have been taken into account in interpreting these experiments. The mass of an alpha particle is 4.00 u (atomic mass units), and its distance of closest approach to the nuclear center in these experiments was about 30 fm. (The wave nature of matter was not postulated until more than a decade after these crucial experiments were first performed.)arrow_forward
- The root mean square speed of the hydrogen molecules at temperature t °C is given by 3x8.31 x (t+273) m 2 x 10-3 Calculate the de Broglie wavelength (in nanometers) of the hydrogen molecules at temperature 24 °C. The mass of the hydrogen molecule is 2 x 1.66 x 10-27 kg. Use two decimals in your answer.arrow_forwardAn 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?arrow_forwardFind the de Broglie wavelength of the thermal neutrons at temperature 27° C. (a) 1.45x100 m (b) 1.45 x10 m (c) 1.45x10-¹² m (d) 1.45x10-¹ marrow_forward
- The diameter of the nucleus is about 12.1fm. What is the kinetic energy of a proton with a de Broglie wavelength of 12.1fm?arrow_forwardWhat must be the velocity, in meters per second, of a beam of electrons if they are to display a de Broglie wavelength of 29.1 μm?arrow_forwardAn x ray tube has an applied voltage of 100 kV. (a) What is the most energetic x-ray photon it can produce? Express your answer in electron volts and joules. (b) Find the wavelength of such an X–ray. (1 MeV = 1.60 x 10-13 J)arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning