An electron has a momentum of 2.00 x 10²¹ kg-m/s. Find the kinetic energy of the electron in units of MeV. (A) 2.31 (B) 5.74 (C) 4.97 (D) 3.27 (E) 6.49
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- An unstable particle with a mass equal to 3.34 ✕ 10−27 kg is initially at rest. The particle decays into two fragments that fly off with velocities of 0.981c and −0.861c, respectively. Find the masses of the fragments in kg. (Hint: Conserve both mass–energy and momentum.)The mass of a proton is 1.67 × 10-27 kg. A. Find the rest energy in joules. B. Find the rest energy in mega-electron volts.The muon is an unstable subatomic particle witha mean lifetime of 2.2 μs, and about 10,000 muons reach everysquare meter of the earth's surface a minute. A muon has a linear momentum G, and after some short time it decays into two other elementary particles with masses m1 and m2. The energy Q released during the decay is converted entirely into the kinetic energy of the decay products (i.e., into kinetic energy of m1 and m2). Find linear momenta of decay products. Draw a vector diagram showing a range of momenta decay products can take.
- 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)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. (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)An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 3.20 ✕ 10−28 kg and that of the heavier fragment is 1.71 ✕ 10−27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment? ?c
- dx 4x 3. In so-called "natural units" (which is just a sneaky way to let us ignore a bunch of constants), the relativistic kinetic energy of a rigid body is given by the formula 1 КЕ — т V1 – v2 where m is the rest mass of the body and v is its relative speed. Alien scientists on a space station are observing an object falling into a black hole. As the object falls, it is disintegrating, losing mass at a rate of 3 (so its mass is changing at a rate of -3). How fast is the kinetic energy of the main part of the object changing when its mass is 20, its velocity is .7, and it is accelerating at a rate of .1 (remember that acceleration is the derivative of velocity with respect to time: a = dt 1Note that this formula does not make sense when v > 1. That is because in natural units, a speed of 1 corresponds to the speed of light, and nothing with positive rest mass can go that fast.Compute the kinetic energy of a proton (mass 1.67 x 10-27 kg) using both the nonrelativis- tic and relativistic expressions (a) 8.0 x 107 m/s and (b) 2.858 m/sA futuristic rocket ship of mass m,hip = 6.0 × 10° kg is sent into space carrying one person with the goal of traveling to nearby stars. The ship uses a newly discovered matter-antimatter drive that converts matter into energy at 100% efficiency. We will take Earth as the rest frame. For parts (a) through (d) you are interested in the portion of the spaceship's travel where it is using its drive to convert matter into the kinetic energy of the ship. (a) What is the speed of the ship when the work done by the ship's drive is equal to 150,000 TWh (the total energy used by humanity in 2014)? (b) How much work must the drive do for the rocket to reach a final speed of 0.95c? (c) How much matter would the drive need to annihilate in order for the ship to reach the final speed of 0.95c? Assume that the matter used in the antimatter drive is collected from the interstellar medium and that the mass of the ship remains constant. (d) When at the final speed of 0.95c, by what factor is the ship's…
- Consider a proton that has a momentum of 1.00 kg·m/s. Part (a) Calculate the relativistic parameter γ for the proton. You will have to use a 2.998 × 108 for the speed of light to get this answer correct. Part (b) What is its speed, in meters per second? Such protons form a rare component of cosmic radiation with uncertain origins.10 L Special Relativity predicts that high-energy electrons increase in mass as they approach the speed of light. What kinetic energy will an electron have if it is travelling at 99.875% of the speed of light? Give your answer in MeV and in J. How many times greater is your result than the value the classical formula mv² gives? G AUAstrophysics 14 Use Newton's second law, F = dp/dt, and the formula for relativistic momentum, Eq. ( 44), to show that the acceleration vector a = dv/dt produced by a force F acting on a particle of mass m is F (F. v), ymc² Ym where F· v is the vector dot product between the force F and the particle velocity v. Thus the acceleration depends on the particle's velocity and is not in general in the same direction as the force. mv p = /1 – v²/c² = ymv (44) VI -