Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Question
Chapter 39, Problem 46AP
(a)
To determine
De Broglie wavelength of neutron having kinetic energy
(b)
To determine
Check whether the answer of part (a) is comparable with the interatomic spacing of crystal.
(c)
To determine
Check whether the thermal neutrons can exhibit diffraction effects in crystal or not.
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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.
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?
An alpha particle (m = 6.64 × 10−27 kg) emitted in the radioactive decay of Uranium-238 has an energy of 4.20 MeV. What is its de Broglie wavelength?
Chapter 39 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 39.1 - Prob. 39.1QQCh. 39.2 - Prob. 39.2QQCh. 39.2 - Prob. 39.3QQCh. 39.2 - Prob. 39.4QQCh. 39.3 - Prob. 39.5QQCh. 39.5 - Prob. 39.6QQCh. 39.6 - Prob. 39.7QQCh. 39 - Prob. 1PCh. 39 - Prob. 2PCh. 39 - Prob. 3P
Ch. 39 - Prob. 4PCh. 39 - Prob. 5PCh. 39 - Prob. 6PCh. 39 - Prob. 8PCh. 39 - Prob. 9PCh. 39 - Prob. 10PCh. 39 - Prob. 11PCh. 39 - Prob. 12PCh. 39 - Prob. 13PCh. 39 - Prob. 15PCh. 39 - Prob. 16PCh. 39 - Prob. 17PCh. 39 - Prob. 18PCh. 39 - Prob. 19PCh. 39 - Prob. 20PCh. 39 - Prob. 22PCh. 39 - Prob. 23PCh. 39 - Prob. 24PCh. 39 - Prob. 25PCh. 39 - Prob. 26PCh. 39 - Prob. 27PCh. 39 - Prob. 30PCh. 39 - Prob. 31PCh. 39 - Prob. 32PCh. 39 - Prob. 33PCh. 39 - Prob. 35PCh. 39 - Prob. 37PCh. 39 - Prob. 38PCh. 39 - Prob. 39PCh. 39 - Prob. 40APCh. 39 - Prob. 41APCh. 39 - Prob. 43APCh. 39 - Prob. 44APCh. 39 - Prob. 45APCh. 39 - Prob. 46APCh. 39 - Prob. 47CPCh. 39 - Prob. 48CPCh. 39 - Prob. 49CPCh. 39 - Prob. 50CP
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- A neutron with a potential energy of U = 0 is represented by the wave function y = Ae'(kx – ot) where k = 5.23 × 1010 m-1. (a) What is the momentum (in kg · m/s) of the neutron? kg • m/s (b) What is the kinetic energy (in eV) of the neutron? ev (c) We wish to verify if the given wave function is a solution to Schrödinger's equation. To do this, calculate the following. (Use the following as necessary: k, w, x, t and A. Do not substitute numerical values; use variables only.) dự dx d?w dx2 Substitute the above two results into Schrödinger's equation and solve for the energy of the particle. (Use the following as necessary: h, k, m, w, x, t and A. Do not substitute numerical values; use variables only.) E = Calculate the energy (in eV) of the neutron. evarrow_forwardSuppose a star with radius 8.50 × 108 m has a peak wavelength of 685 nm in the spectrum of its emitted radiation. (a) Find the energy of a photon with this wavelength. (b) What is the surface temperature of the star? (c) At what rate is energy emitted from the star in the form of radiation? Assume the star is a blackbody (e = 1). (d) Using the answer to part (a), estimate the rate at which photons leave the surface of the star.arrow_forwardThe 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_forward
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