EBK PHYSICAL UNIVERSE
15th Edition
ISBN: 9780100255036
Author: KRAUSKOPF
Publisher: YUZU
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 9, Problem 24E
To determine
The experiment used to differentiate an electron and gamma ray of same wavelength.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
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?
Find 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 m
At what speed in units of m/s, will an electron have a de Broglie wavelength of 41.4 μm?
Chapter 9 Solutions
EBK PHYSICAL UNIVERSE
Ch. 9 - Prob. 1MCCh. 9 - Prob. 2MCCh. 9 - Prob. 3MCCh. 9 - Prob. 4MCCh. 9 - Prob. 5MCCh. 9 - Prob. 6MCCh. 9 - Prob. 7MCCh. 9 - Prob. 8MCCh. 9 - Prob. 9MCCh. 9 - Prob. 10MC
Ch. 9 - Prob. 11MCCh. 9 - Prob. 12MCCh. 9 - Prob. 13MCCh. 9 - Prob. 14MCCh. 9 - Prob. 15MCCh. 9 - Prob. 16MCCh. 9 - Prob. 17MCCh. 9 - Prob. 18MCCh. 9 - Prob. 19MCCh. 9 - Prob. 20MCCh. 9 - Prob. 21MCCh. 9 - Prob. 22MCCh. 9 - Prob. 23MCCh. 9 - Prob. 24MCCh. 9 - Prob. 25MCCh. 9 - Prob. 26MCCh. 9 - Prob. 27MCCh. 9 - Prob. 28MCCh. 9 - Prob. 29MCCh. 9 - Prob. 30MCCh. 9 - Prob. 31MCCh. 9 - Prob. 32MCCh. 9 - Prob. 33MCCh. 9 - Prob. 34MCCh. 9 - Prob. 35MCCh. 9 - Prob. 36MCCh. 9 - Prob. 37MCCh. 9 - Prob. 38MCCh. 9 - Prob. 39MCCh. 9 - Prob. 40MCCh. 9 - Prob. 41MCCh. 9 - Prob. 42MCCh. 9 - Prob. 1ECh. 9 - Prob. 2ECh. 9 - Prob. 3ECh. 9 - Prob. 4ECh. 9 - Prob. 5ECh. 9 - Prob. 6ECh. 9 - Prob. 7ECh. 9 - Prob. 8ECh. 9 - Prob. 9ECh. 9 - Prob. 10ECh. 9 - Prob. 11ECh. 9 - Prob. 12ECh. 9 - Prob. 13ECh. 9 - Prob. 14ECh. 9 - Prob. 15ECh. 9 - Prob. 16ECh. 9 - Prob. 17ECh. 9 - Prob. 18ECh. 9 - Prob. 19ECh. 9 - Prob. 20ECh. 9 - Prob. 21ECh. 9 - Prob. 22ECh. 9 - Prob. 23ECh. 9 - Prob. 24ECh. 9 - Prob. 25ECh. 9 - Prob. 26ECh. 9 - Prob. 27ECh. 9 - Prob. 28ECh. 9 - Prob. 29ECh. 9 - Prob. 30ECh. 9 - Prob. 31ECh. 9 - Prob. 32ECh. 9 - Prob. 33ECh. 9 - Prob. 34ECh. 9 - Prob. 35ECh. 9 - Prob. 36ECh. 9 - Prob. 37ECh. 9 - Prob. 38ECh. 9 - Prob. 39ECh. 9 - Prob. 40ECh. 9 - Prob. 41ECh. 9 - Prob. 42ECh. 9 - Prob. 43ECh. 9 - Prob. 44ECh. 9 - Prob. 45ECh. 9 - Prob. 46ECh. 9 - Prob. 47ECh. 9 - Prob. 48ECh. 9 - Prob. 49ECh. 9 - Prob. 50ECh. 9 - Prob. 51ECh. 9 - Prob. 52ECh. 9 - Prob. 53ECh. 9 - Prob. 54ECh. 9 - Prob. 55ECh. 9 - Prob. 56ECh. 9 - Prob. 57ECh. 9 - Prob. 58ECh. 9 - Prob. 59ECh. 9 - Prob. 60ECh. 9 - Prob. 61ECh. 9 - Prob. 62ECh. 9 - Under what circumstances do electrons exhibit...
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
- What is the de Brogue wavelength of a proton whose kinetic energy is 2.0 MeV? 10.0 MeV?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_forwardThe 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.arrow_forward
- If the de Broglie wavelength of a neutron is 10 -10m what is the energy associated with it? Mass of neutron is 1.675×10 -27 kg.arrow_forwardWhat is the de Broglie wavelength of a proton whose kinetic energy is 2.0 MeV?arrow_forwardWhat is the de Broglie wavelength of a proton whose kinetic energy is 2.0 MeV? 10.0 MeV?arrow_forward
- What is the de Broglie wavelength of a electron that is moving at 8.77 x 105 m/s? Please give your answer in nanometers.arrow_forwardPhotoelectrons with a maximum kinetic energy of 7.95 eV are emitted from a metal when it is illuminated by ultraviolet radiation wavelength of 1.25 × 10-1 m. What is the de Broglie wavelength (in nanometres) of an electron with a kinetic energy of 7.95 eV?arrow_forwardA) 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.) B) Calculate the velocity of an electron (me = 9.10939×10-31 kg) having a de Broglie wavelength of 230.1 pm.arrow_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_forwardThe energy conservation principle that applies to the photoelectric experiment is Ephoton = (KE)electron + W, where W is the “work function” for the metal. (The work function is the minimum energy required to eject an electron from the metal surface.) The work function for calcium metal is 4.60 10−19 J. If calcium is irradiated with 400-nm photons, what is the de Broglie wavelength of the resulting photoelectron beam?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
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 LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning