University Physics with Modern Physics, Books a la Carte Edition; Modified MasteringPhysics with Pearson eText -- ValuePack Access Card -- for ... eText -- Valuepack Access Card (14th Edition)
14th Edition
ISBN: 9780134308142
Author: Hugh D. Young, Roger A. Freedman
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 39, Problem 39.7DQ
To determine
Why recoil energy term is neglected from the energy equation for an atom.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
In solid helium the spacing between atoms is about 3Å . Helium contains 2 protons and 2 neutrons, so the mass of a helium atom is 6.6×10−27 kg. What is the smallest possible energy of a helium atom in solid helium? What temperature (in Kelvin) does this energy correspond with? (Boltzmann's constant is kb=1.38×10−23 J/K
X-rays of wavelength λλ = 0.140 nmnm are scattered from carbon. What is the Compton wavelength shift for photons detected at angle (relative to the incident beam) of exactly 30.0 ∘∘?What is the Compton wavelength shift for photons detected at angle (relative to the incident beam) of exactly 120 ∘∘? What is the Compton wavelength shift for photons detected at angle (relative to the incident beam) of exactly 180 ∘∘?
A parallel beam of 500-keV photons is normally incident on 0.8 cm sheet of lead (density = 11.4 g/cm³) at a
rate of 2x10 photns/sec. µ/p = 0.16 cm²/g, Hen/p = 0.09 cm²/g, H/p = 0.11 cm²/g
What is the fraction of photons transmitted without interaction?
Choose. +
What fraction of the transmitted energy is due to uncollided photons?
Choose...
What fraction of the initial kinetic energy transferred to the electrons is emitted as
Bremsstrahlung?
Choose...
Chapter 39 Solutions
University Physics with Modern Physics, Books a la Carte Edition; Modified MasteringPhysics with Pearson eText -- ValuePack Access Card -- for ... eText -- Valuepack Access Card (14th Edition)
Ch. 39.2 - Prob. 39.2TYUCh. 39.3 - Prob. 39.3TYUCh. 39.4 - Prob. 39.4TYUCh. 39.5 - Prob. 39.5TYUCh. 39.6 - Prob. 39.6TYUCh. 39 - Prob. 39.1DQCh. 39 - Prob. 39.2DQCh. 39 - Prob. 39.3DQCh. 39 - When an electron beam goes through a very small...Ch. 39 - Prob. 39.5DQ
Ch. 39 - Prob. 39.6DQCh. 39 - Prob. 39.7DQCh. 39 - Prob. 39.8DQCh. 39 - Prob. 39.9DQCh. 39 - Prob. 39.10DQCh. 39 - Prob. 39.11DQCh. 39 - Prob. 39.12DQCh. 39 - Prob. 39.13DQCh. 39 - Prob. 39.14DQCh. 39 - Prob. 39.15DQCh. 39 - Prob. 39.16DQCh. 39 - Prob. 39.17DQCh. 39 - Prob. 39.18DQCh. 39 - Prob. 39.19DQCh. 39 - Prob. 39.20DQCh. 39 - Prob. 39.21DQCh. 39 - When you check the air pressure in a tire, a...Ch. 39 - Prob. 39.1ECh. 39 - Prob. 39.2ECh. 39 - Prob. 39.3ECh. 39 - Prob. 39.4ECh. 39 - Prob. 39.5ECh. 39 - Prob. 39.6ECh. 39 - Prob. 39.7ECh. 39 - Prob. 39.8ECh. 39 - Prob. 39.9ECh. 39 - Prob. 39.10ECh. 39 - Prob. 39.11ECh. 39 - Prob. 39.12ECh. 39 - Prob. 39.13ECh. 39 - Prob. 39.14ECh. 39 - Prob. 39.15ECh. 39 - Prob. 39.16ECh. 39 - Prob. 39.17ECh. 39 - Prob. 39.18ECh. 39 - Prob. 39.19ECh. 39 - Prob. 39.20ECh. 39 - Prob. 39.21ECh. 39 - Prob. 39.22ECh. 39 - Prob. 39.23ECh. 39 - Prob. 39.24ECh. 39 - Prob. 39.25ECh. 39 - Prob. 39.26ECh. 39 - Prob. 39.27ECh. 39 - Prob. 39.28ECh. 39 - Prob. 39.29ECh. 39 - Prob. 39.30ECh. 39 - Prob. 39.31ECh. 39 - Prob. 39.32ECh. 39 - Prob. 39.33ECh. 39 - Prob. 39.34ECh. 39 - Prob. 39.35ECh. 39 - Prob. 39.36ECh. 39 - Prob. 39.37ECh. 39 - Prob. 39.38ECh. 39 - Prob. 39.39ECh. 39 - Prob. 39.40ECh. 39 - Prob. 39.41ECh. 39 - Prob. 39.42ECh. 39 - Prob. 39.43ECh. 39 - Prob. 39.44ECh. 39 - Prob. 39.45ECh. 39 - Prob. 39.46ECh. 39 - Prob. 39.47ECh. 39 - Prob. 39.48ECh. 39 - Prob. 39.49ECh. 39 - Prob. 39.50PCh. 39 - Prob. 39.51PCh. 39 - Prob. 39.52PCh. 39 - Prob. 39.53PCh. 39 - Prob. 39.54PCh. 39 - Prob. 39.55PCh. 39 - Prob. 39.56PCh. 39 - Prob. 39.57PCh. 39 - Prob. 39.58PCh. 39 - Prob. 39.59PCh. 39 - An Ideal Blackbody. A large cavity that has a very...Ch. 39 - Prob. 39.61PCh. 39 - Prob. 39.62PCh. 39 - Prob. 39.63PCh. 39 - Prob. 39.64PCh. 39 - Prob. 39.65PCh. 39 - Prob. 39.66PCh. 39 - Prob. 39.67PCh. 39 - Prob. 39.68PCh. 39 - Prob. 39.69PCh. 39 - Prob. 39.70PCh. 39 - Prob. 39.71PCh. 39 - Prob. 39.72PCh. 39 - Prob. 39.73PCh. 39 - Prob. 39.74PCh. 39 - Prob. 39.75PCh. 39 - Prob. 39.76PCh. 39 - Prob. 39.77PCh. 39 - Prob. 39.78PCh. 39 - Prob. 39.79PCh. 39 - Prob. 39.80PCh. 39 - A particle with mass m moves in a potential U(x) =...Ch. 39 - Prob. 39.82PCh. 39 - Prob. 39.83PCh. 39 - DATA In the crystallography lab where you work,...Ch. 39 - Prob. 39.85PCh. 39 - Prob. 39.86CPCh. 39 - Prob. 39.87CPCh. 39 - Prob. 39.88PPCh. 39 - Prob. 39.89PPCh. 39 - Prob. 39.90PPCh. 39 - Prob. 39.91PP
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
- A) Calculate the de Broglie wavelength of a neutron (mn = 1.67493×10-27 kg) moving at one six hundredth of the speed of light (c/600). Enter at least 4 significant figures. (I got the answer 949.4 pm but it is wrong, please help) B) Calculate the velocity of an electron (me = 9.10939×10-31 kg) having a de Broglie wavelength of 230.1 pm.arrow_forwardAn electron was ejected from the surface of cesium metal with a speed of 6.74*10^5 m/s.Given that the threshold energy for the removal of an electron from the cesium metal is 2.90*10^-19J. Calculate the incident wavelength in nanometers. For this problem, I believe we use the KE=E photon-Eo. However, I am having trouble getting from the formula to the desired answer of 400 nm.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
- Calculate the following energies: the kinetic energy of a lead K-shell photoelectron following a photoelectric interaction of a 662 keV gamma ray (K-shell binding energy of Pb = 88 keV).arrow_forwardShow that the maximum kinetic energy of the recoil electron in Compton scattering is given by 2hf mc² K. E.max (electron) = hf · 2hf 1+ mc² At what angles and does this occur? If we detect a scattered electron at angle = 0° of 100 keV, what energy photon was scattered?arrow_forwardThe two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1 and T2, respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B?arrow_forward
- answer only 4arrow_forwardMost microwave ovens emit electro-magnetic radiation with a wavelength () of 12.24 cm. This EM-radiation is used to heat stuff up, like food, and it is emitted at a rate of 6.30 × 1026 photons/second for a typical microwave. How long will it take to boil a 175.0 mL cup of water initially at 25.5 ºC (water boils at 100.0 ºC) if 45.0% of the photons emitted by the microwave oven are absorbed by the water? My answer does not even make any sort of rational sense...arrow_forward(a) Find the momentum (in kg - m/s) of a 44.5 kev x-ray photon. |kg · m/s (b) Find the equivalent velocity (in m/s) of a neutron with the same momentum. m/s (c) What is the neutron's kinetic energy (in ev)? evarrow_forward
- In a Compton scattering experiment, an x-ray photon scatters through an angle of 15.0° from a free electron that is initially at rest. The electron recoils with a speed of 2,200 km/s. (a) Calculate the wavelength of the incident photon. nm (b) Calculate the angle through which the electron scatters. Oarrow_forwardGold has a work function, Φ, of 5.47 ev. If a sample of gold is bombarded with 175 nm photons, what will be the velocity (in m/s) of the photo-ejected electrons? (1 eV = 1.60 × 10⁻¹⁹ J, the mass of an electron is 9.109 x 10⁻³¹ kg, and Planck's constant is 6.626 x 10⁻³⁴ Js.)arrow_forwardA 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_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning