Modern Physics
3rd Edition
ISBN: 9781111794378
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
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Chapter 3, Problem 47P
To determine
The expression for the kinetic energy for the electron after a head-on collision with a photon.
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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.
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.
An 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?
Chapter 3 Solutions
Modern Physics
Ch. 3.2 - Calculate the quantum number, n, for this pendulum...Ch. 3.2 - An object of mass m on a spring of stiffness k...Ch. 3 - Prob. 1QCh. 3 - Prob. 2QCh. 3 - Prob. 3QCh. 3 - Prob. 4QCh. 3 - Prob. 5QCh. 3 - Prob. 6QCh. 3 - Prob. 7QCh. 3 - Prob. 8Q
Ch. 3 - Prob. 9QCh. 3 - Prob. 10QCh. 3 - Prob. 11QCh. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10PCh. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 19PCh. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - As a single crystal is rotated in an x-ray...Ch. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 44PCh. 3 - Prob. 46PCh. 3 - Prob. 47PCh. 3 - Prob. 48P
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- 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_forwardHow fast would a proton have to be traveling in order to have the same momentum as a photon with a frequency of 8.18 x 10^18 Hz?arrow_forwardElectron capture is a variant on beta-radiation. The lightest nucleus to decay by electron capture is 7Be -- beryllium-7. The daughter nucleus is 7Li -- lithium-7. The electron is transformed into a massless particle (a neutrino): e − + 7 B e + ⟶ 7 L i + ν The initial electron is bound in the atom, so the beryllium mass includes the electron. In fact, since the electron starts bound in the atom, a more-accurate statement of the nuclear reaction is probably: 7 B e ⟶ 7 L i + ν The masses are beryllium: 7.016929 u, and lithium: 7.016003 u, and refer to the neutral atom as a whole. (Use uc and uc2 as your momentum and energy units -- but carry them along in your calculation.) The initial beryllium atom is stationary. Calculate the speed of the final lithium nucleus in km/s. (You will make life much easier for yourself if you recognize that practically all the energy released goes into the lighter particle. c = 300,000 km/s)arrow_forward
- Electron capture is a variant on beta-radiation. The lightest nucleus to decay by electron capture is 7Be -- beryllium-7. The daughter nucleus is 7Li -- lithium-7. The electron is transformed into a massless particle (a neutrino): e − + 7 B e + ⟶ 7 L i + ν The initial electron is bound in the atom, so the beryllium mass includes the electron. In fact, since the electron starts bound in the atom, a more-accurate statement of the nuclear reaction is probably: 7 B e ⟶ 7 L i + ν The masses are beryllium: 7.016929 u, and lithium: 7.016003 u, and refer to the neutral atom as a whole. (Use uc and uc2 as your momentum and energy units -- but carry them along in your calculation.) The initial beryllium atom is stationary. Calculate the speed of the final lithium nucleus in km/s. (all the energy released goes into the lighter particle. c = 300,000 km/s)arrow_forwardA photon of frequency v is scattered by an electron initially at rest. Verify that the maximum kinetic energy of the recoil electron is KEmax = (2h² v/mc²)/(1 + 2 hv/mc²).arrow_forwardAn electron traveling with a speed of 2.5 x 10^6 meters per second collides with a photon having a frequency of 1 x 10^16 hertz. After the collision, the photon has 3.18 x 10^-18 joule of energy. Determine the energy in joules of the photon before the collision.arrow_forward
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