Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 31, Problem 4P
(a)
To determine
The magnitude of minimum frequency.
(b)
To determine
The wavelength of each photon.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
A linear particle accelerator using beta particles collides electrons with their anti-matter counterparts, positrons. The accelerated electron hits the stationary positron with a velocity of 19 x 106 m/s, causing the two particles to annihilate.If two gamma photons are created as a result, calculate the energy of each of these two photons, giving your answer in MeV (mega electron volts), accurate to 1 decimal place. Take the mass of the electron to be 5.486 x 10-4 u, or 9.109 x 10-31 kg.Note: Assume that the kinetic energy is also converted into the gamma rays, and is included in the two photons.
An electron and a positron are moving toward each other and each has speed 0.500c in the lab frame. (a) What is the kinetic energy of each particle? (b) The e+ and e- meet head-on and annihilate. What is the energy of each photon that is produced? (c) What is the wavelength of each photon? How does the wavelength compare to the photon wavelength when the initial kinetic energy of the e+ and e- is negligibly small.
A linear particle accelerator using beta particles collides electrons with their anti-matter counterparts, positrons. The accelerated electron hits the stationary positron with a velocity of 29 x 106 m/s, causing the two particles to annihilate.If two gamma photons are created as a result, calculate the energy of each of these two photons, giving your answer in MeV (mega electron volts), accurate to 1 decimal place. Take the mass of the electron to be 5.486 x 10-4 u, or 9.109 x 10-31 kg.
Chapter 31 Solutions
Principles of Physics: A Calculus-Based Text
Ch. 31.2 - Given the identification of the particles in...Ch. 31.5 - Prob. 31.2QQCh. 31.5 - Prob. 31.3QQCh. 31.5 - Prob. 31.4QQCh. 31.9 - Prob. 31.5QQCh. 31 - Prob. 1OQCh. 31 - Prob. 2OQCh. 31 - Prob. 3OQCh. 31 - Prob. 4OQCh. 31 - Prob. 5OQ
Ch. 31 - Prob. 6OQCh. 31 - Prob. 7OQCh. 31 - Prob. 8OQCh. 31 - Prob. 1CQCh. 31 - Prob. 2CQCh. 31 - Prob. 3CQCh. 31 - Prob. 4CQCh. 31 - Prob. 5CQCh. 31 - Prob. 6CQCh. 31 - Prob. 7CQCh. 31 - Prob. 8CQCh. 31 - Prob. 9CQCh. 31 - Prob. 10CQCh. 31 - Prob. 11CQCh. 31 - Prob. 12CQCh. 31 - Prob. 13CQCh. 31 - Prob. 1PCh. 31 - Prob. 2PCh. 31 - Prob. 3PCh. 31 - Prob. 4PCh. 31 - Prob. 5PCh. 31 - Prob. 6PCh. 31 - Prob. 7PCh. 31 - Prob. 8PCh. 31 - Prob. 9PCh. 31 - Prob. 10PCh. 31 - Prob. 11PCh. 31 - Prob. 12PCh. 31 - Prob. 13PCh. 31 - Prob. 14PCh. 31 - Prob. 15PCh. 31 - Prob. 16PCh. 31 - Prob. 17PCh. 31 - Prob. 18PCh. 31 - Prob. 19PCh. 31 - Prob. 20PCh. 31 - Prob. 21PCh. 31 - Prob. 22PCh. 31 - Prob. 23PCh. 31 - Prob. 24PCh. 31 - Prob. 25PCh. 31 - Prob. 26PCh. 31 - Prob. 27PCh. 31 - Prob. 28PCh. 31 - Prob. 29PCh. 31 - Prob. 30PCh. 31 - Prob. 31PCh. 31 - Prob. 32PCh. 31 - Prob. 33PCh. 31 - Prob. 34PCh. 31 - Prob. 35PCh. 31 - Prob. 36PCh. 31 - Prob. 37PCh. 31 - Prob. 38PCh. 31 - Prob. 39PCh. 31 - Prob. 40PCh. 31 - Prob. 41PCh. 31 - Prob. 42PCh. 31 - Prob. 43PCh. 31 - Prob. 44PCh. 31 - Prob. 45PCh. 31 - Prob. 46PCh. 31 - Prob. 47PCh. 31 - Prob. 48PCh. 31 - Prob. 49PCh. 31 - Prob. 50PCh. 31 - Prob. 51PCh. 31 - Prob. 52PCh. 31 - Prob. 53PCh. 31 - Prob. 54PCh. 31 - Prob. 55PCh. 31 - Prob. 56PCh. 31 - Prob. 57PCh. 31 - Prob. 58PCh. 31 - Prob. 59PCh. 31 - Prob. 60PCh. 31 - Prob. 61PCh. 31 - Prob. 62PCh. 31 - Prob. 63PCh. 31 - Prob. 64PCh. 31 - Prob. 65P
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
- An electron and a positron, initially far apart, move toward each other with the same speed. They collide head-on, annihilating each other and producing two photons. Find the energies, wavelengths, and frequencies of the photons if the initial kinetic energies of the electron and positron are (a) both negligible and (b) both 5.000 MeV. The rest energy of an electron or a positron is 0.511 MeV.arrow_forwardAn electron and a positron anihilate with equal and opposite momenta and become two photons: e- +e- → 2 photons. Both the electron and the positron have mass me and momentum of magnitude mec. Explain why the electron and positron cannot annihilate to become a single (a) photon. (b) What are the magnitudes of the momenta of the photons?arrow_forwardAs a scientist in a nuclear physics research lab, you are conducting a photodisintegration experiment to verify the binding energy of a deuteron. A photon with wavelength l in air is absorbed by a deuteron, which breaks apart into a neutron and a proton. The two fragments share the released kinetic energy equally, and the deuteron can be assumed to be initially at rest. You measure the speed of the proton after the disintegration as a function of the wavelength l of the photon. Your experimental results are given in the table . (a) Graph the data as y2 versus 1/l. Explain why the data points, when graphed this way, should follow close to a straight line. Find the slope and y-intercept of the straight line that gives the best fit to the data. (b) Assume that h and c have their accepted values. Use your results from part (a) to calculate the mass of the proton and the binding energy (in MeV) of the deuteron.arrow_forward
- A muon is a short-lived particle. Muons are created by cosmic rays; they can also be created by particle accelerators. The muon is similar to an electron but has a larger mass: mμ ≈ 200me. During its brief lifetime, a muon can combine with a proton to create a system that is similar to atomic hydrogen called a muonic hydrogen atom. The larger mass of the muon makes some of the assumptions of the Bohr hydrogen atom treatment less accurate, but using the mathematics of the Bohr hydrogen atom to analyze this system will give approximate results that allow us to understand how the changing mass affects the properties of the system. How does the energy required to ionize a muonic hydrogen atom compare to that required to ionize a regular hydrogen atom?A. It is greater.B. It is approximately the same.C. It is less.arrow_forwardAn antiproton has the same mass and rest energy (938.3 MeV) as a proton, but has a negative change -e instead of the positive charge +e of the proton. A proton and an antiproton, initially far apart, move toward each other with the same speed and collide head-on, annihilating each other and producing two photons. Find the energies and wavelengths of the photons if the initial kinetic energies of the proton and antiproton are (a) both negligible and (b) both 545 MeV.arrow_forwardA muon is a short-lived particle. Muons are created by cosmic rays; they can also be created by particle accelerators. The muon is similar to an electron but has a larger mass: mμ ≈ 200me. During its brief lifetime, a muon can combine with a proton to create a system that is similar to atomic hydrogen called a muonic hydrogen atom. The larger mass of the muon makes some of the assumptions of the Bohr hydrogen atom treatment less accurate, but using the mathematics of the Bohr hydrogen atom to analyze this system will give approximate results that allow us to understand how the changing mass affects the properties of the system. The larger mass of the muon complicates an accurate mathematical treatment similar to that of the Bohr hydrogen atom becauseA. The de Broglie wavelength of the muon is shorter than that of the electron.B. The relatively small difference in mass between the muon and the proton means that we can’t ignore the motion of the proton.C. The short lifetime of the muon…arrow_forward
- An atom of mass m₁ = m moves in the positive x direc- tion with speed v₁ = v. It collides with and sticks to an atom of mass m₂ = 2m moving in the positive y direction with speed v₂ = 2v/3. Find the resultant speed and direc- tion of motion of the combination, and find the kinetic energy lost in this inelastic collision.arrow_forwardThe least massive particle known to exist is the electron neutrino. Though scientists once believed that it had no mass, like the photon, they have now determined that this particle has an extremely low mass, equivalent to a few electron volts. Assuming a mass of 2.2 eV/c2 (or 3.9 × 10-36 kg) and a speed of 4.4 × 106 m/s, which of the following values equals the neutrino’s de Broglie wavelength? a. 3.8 × 10-5 m b. 4.7 × 10-7m c. 1.7 × 10-10m d. 8.9 × 10-14marrow_forwardThe muon is unstable and has a mean lifetime of about 2.2 microseconds. A muon is an elementary particle similar to the electron. with an electric charge of -1 e, but with a much greater mass. The mass of a muon is 0.1135u, where the Atomic mass unit u = 1.66 x 10^-27 kg. It decays into an electron and two neutrinos. What is the energy released in this decay (in MeV)? Recall 1 eV = 1.6 x 10^-19 J).arrow_forward
- (a) Show that the kinetic energy of a nonrelativistic particle can be written in terms of its momentum as KE =p2/2m. (b) Use the results of part (a) to find the minimum kinetic energy of a proton confined within a nucleus having a diameter of 1.0 × 10−15m.arrow_forwardAn unknown moving ion is confined in a OD nanomaterial in which all three dimensions are equals to 5 nm. Estimate with what accuracy its velocity and energy can be measured (given mass of the ion is 4.8×10 26 kg)?arrow_forwardA particle and its anti-particle have the same mass and charge, but the charges will be of opposite polarity. A particle and its anti-particle annihilate and produce two photons. The particle has kinetic energy 2.0 MeV collides head on with its anti-particle that is at rest. Find the energy and momentum of each photon. (Given the mass of each particle is 0.632MeV/c²) Select the correct answer choice: (a) E = 1.088 MeV E = 3.264 MeV (b) (c) (d) E = 1.632 MeV E = 1.088 MeV p=1.67 MeV/c p=1.183 MeV/c p=2.366 MeV/c p=0.835 MeV/carrow_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
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning