Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 45, Problem 34P
(a)
To determine
The half thickness for the material.
(b)
To determine
The thickness changes of the material by a factor of f.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
In a diagnostic x-ray procedure, 5.25×1010 photons are absorbed by tissue with a mass of 0.590 kg. The x-ray wavelength is 2.00×10−2 nm.
A) What is the total energy absorbed by the tissue?
Express your answer in millijoules.
B) What is the equivalent dose in rem?
Express your answer in rem.
A small 10-gram source of cobalt-60 is in a vacuum. (a) What is the activity of the cobalt-60source in Bq? (b) What is the actual gamma-ray flux in cm2-s-1 at a point of measurement500 cm from the source due to the cobalt-60 emitted gamma-rays? (c) If the backgroundgamma-ray flux in the vicinity of the 10-g source is 7.8×107cm-2-s-1, what is the totalgamma-ray flux at the point of measurement, including the background? (d) What thicknessof a lead shield in cm would have to be placed between the source and the point ofmeasurement to reduce the total of the background plus the uncollided gamma-ray flux fromthe cobalt-60 source to 1.00×108cm-2-s-1 at that point?
(I've attempted part a of the problem and don't know how to continue)
The intensity of radioactivity depends on the distance of the recipient from the source of radiation and follows an inverse square law. Suppose an object receives 1230 millirem of radiation when it is 10 meters from the source, calculate the distance if the object is now exposed to 350 millirem.
Chapter 45 Solutions
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Ch. 45.1 - When a nucleus undergoes fission, the two daughter...Ch. 45.2 - Prob. 45.2QQCh. 45.3 - Prob. 45.3QQCh. 45.4 - Prob. 45.4QQCh. 45 - Prob. 1OQCh. 45 - Prob. 2OQCh. 45 - Prob. 3OQCh. 45 - Prob. 4OQCh. 45 - Prob. 5OQCh. 45 - Prob. 6OQ
Ch. 45 - Prob. 7OQCh. 45 - Prob. 8OQCh. 45 - Prob. 9OQCh. 45 - Prob. 1CQCh. 45 - Prob. 2CQCh. 45 - Prob. 3CQCh. 45 - Prob. 4CQCh. 45 - Prob. 5CQCh. 45 - Prob. 6CQCh. 45 - Prob. 7CQCh. 45 - Prob. 8CQCh. 45 - Prob. 1PCh. 45 - Prob. 2PCh. 45 - Prob. 3PCh. 45 - Prob. 4PCh. 45 - Prob. 5PCh. 45 - Prob. 6PCh. 45 - Prob. 7PCh. 45 - Prob. 8PCh. 45 - Prob. 9PCh. 45 - Prob. 10PCh. 45 - Prob. 11PCh. 45 - Prob. 12PCh. 45 - Prob. 13PCh. 45 - Prob. 14PCh. 45 - Prob. 15PCh. 45 - Prob. 16PCh. 45 - Prob. 18PCh. 45 - Prob. 19PCh. 45 - Prob. 20PCh. 45 - Prob. 21PCh. 45 - Prob. 22PCh. 45 - Prob. 23PCh. 45 - Prob. 24PCh. 45 - Prob. 25PCh. 45 - Prob. 26PCh. 45 - Prob. 27PCh. 45 - Prob. 28PCh. 45 - Prob. 29PCh. 45 - Prob. 30PCh. 45 - Prob. 31PCh. 45 - Prob. 32PCh. 45 - Prob. 33PCh. 45 - Prob. 34PCh. 45 - Prob. 35PCh. 45 - Prob. 36PCh. 45 - Prob. 37PCh. 45 - Prob. 41PCh. 45 - Prob. 42PCh. 45 - Prob. 43PCh. 45 - Prob. 44PCh. 45 - Prob. 45PCh. 45 - Prob. 46APCh. 45 - Prob. 47APCh. 45 - Prob. 48APCh. 45 - Prob. 49APCh. 45 - Prob. 51APCh. 45 - Prob. 52APCh. 45 - Prob. 53APCh. 45 - Prob. 54APCh. 45 - Prob. 55APCh. 45 - Prob. 56APCh. 45 - Prob. 57APCh. 45 - Prob. 58APCh. 45 - Prob. 59APCh. 45 - Prob. 60APCh. 45 - Prob. 61APCh. 45 - Prob. 62APCh. 45 - Prob. 63APCh. 45 - Prob. 64APCh. 45 - Prob. 65APCh. 45 - Prob. 66APCh. 45 - Prob. 67APCh. 45 - Prob. 68APCh. 45 - Prob. 69APCh. 45 - Prob. 70APCh. 45 - Prob. 71APCh. 45 - Prob. 72APCh. 45 - Prob. 73AP
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
- Calculate the dose in Sv to the chest at a patient given an xray under the following conditions. The xray beam intensity is 1.50 W/m2, the area of the chest exposed is 0.0750 m2 35.0% of the xrays are absorbed in 20.0 kg of tissue, and the exposure time is 0.250 s.arrow_forwardWhat is the dose in mSv for: (a) a 0.1 Gy xray? (b) 2.5 mGy of neutron exposure to the eye? (c) 1.5 mGy of exposure?arrow_forwardA 70-kg researcher absorbs 4.5 × neutrons in a work day, each of energy 1.2 MeV. The relative biological effectiveness (RBE) for these neutrons is 10. What is the equivalent dosage of the radiation exposure for this researcher, in millirem?arrow_forward
- During an X-ray examination, a person is exposed to radiation at a rate of 2.80×10-5 grays per second. The exposure time is 0.140 s, and the mass of the exposed tissue is 1.60 kg. Determine the energy absorbed.arrow_forwardDuring X-ray diagnostics, lead aprons made of a thin layer of lead are used in order to protect against the harmful effects of X-ray radiation. Determine, by what factor is the intensity of X-ray radiation reduced after passing the apron of thickness 0.5 mm. The mass attenuation coefficient for a radiation source with an energy of 140 keV is μm = 2 cm²/g. Density of lead ρ = 11.35 g/cm³arrow_forwardLead (density 11.34 g/cm³) and concrete (density 2.4 g/cm³) are highly effective materias in radiation shielding. a) What thickness of lead in cm is required to reduce the intensity of a 0.2 MeV gamma ray beam by a factor of 1000. b) What is the thickness of concrete in cm equivalent to that thickness of lead?arrow_forward
- The linear absorption coefficient for 1-MeV gamma rays in lead is 78 m1. Find the thickness of lead required to reduce by one-fourth the intensity of beam of such gamma rays.arrow_forwardGamma-ray detectors like the one described in the preceding problem often use calorimetry to determine gamma-ray energies. Suppose a beam of 100-MeV gamma rays strikes a target with a mass of 2.5 kg and specifi c heat 430 J/(kg # K). How many gamma rays are needed to raise the target’s temperature by 10 mK?arrow_forwardIf a patient undergoing radiation treatment for prostate cancer receives 41 Gray ( Gy G y ) of gamma radiation, how many rad ( D D ) does this correspond to? (1Gy=1×102D) ( 1 G y = 1 × 10 2 D ) .arrow_forward
- question 4 pleasearrow_forwardGamma rays of energy E, =14.4 keV are emitted when "Fe nuclei make trapsition from the first excited state to the ground state. The lifetime ofthis excited state is about 3x10's. If the spread in the energy of the excited state is written as AE =-E, find the value of v. Here c is the speed of light in vacuum.arrow_forwardB please, helparrow_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 LearningModern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College
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
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College