Physics 111 Homework Solutions Week #9 - Thursday
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1 Physics 111 Homework Solutions Week #9 - Thursday Monday, March 1, 2010 Chapter Based on special relativity we know that as a particle with mass travels near the speed of light its mass increases In order to accelerate this object from rest to a speed near that of light would require an ever increasing force (one that rapidly becomes larger by a factor of γ) There are no known forces that could accelerate a particle with mass to the speed of light in a finite amount of time with a finite amount of energy So for objects with no rest mass, as they travel at the speed of light, there mass does not increase with increasing speed we avoid these problems of accelerating the massless particles 246 The Compton shift in wavelength for the proton the electron are given by Δλ e = h 1 cosφ m e c ( ) Δλ p = h ( 1 cosφ)respectively Evaluating the ratio c of the shift in wavelength for the proton to the electron, evaluated at the same detection angle φ, we find Δλ p = hm c e Δλ e ch = m e = Δλ p = Δλ e Therefore the shift in wavelength for the proton is smaller than the wavelength shift for the electron Multiple-Choice Problems Relativistic energy momentum for an object of mass m For an object with a m = 1kg rest mass, it has a rest energy of E = mc 2 = 9x10 16 J The Lorentz factor is given by:, with the relativistic momentu = γmv, the relativistic energy E 2 = p 2 c 2 +m 2 c 4 a For a velocity of 08c, The relativistic momentum energy are therefore,
2 b Following the procedure in part a, for a velocity of 09c,, the relativistic momentum is 619x10 8 kgm/s, the relativistic energy is 207x10 17 J c For a velocity of 095c, γ = 320, the relativistic momentum is 913x10 8 kgm/s, the relativistic energy is 288x10 17 J d For a velocity of 099c, γ = 709, the relativistic momentum is 211x10 9 kgm/s, the relativistic energy is 6387x10 17 J e For a velocity of 0999c, γ = 224, the relativistic momentum is 670x10 9 kgm/s, the relativistic energy is 20x10 18 J 242 For an electron with rest mass 911x10-31 kg, it has a rest energy of E = mc 2 = 8199x10-14 J = 0511 MeV The Lorentz factor is given by:, the relativistic momentum energy are given respectively as p = γmv a For the electron with a velocity of 08c, γ = 167 The relativistic momentum energy are therefore, Using the fact that 16x10-19 J = 1 ev, the relativistic energy is given as Further it can be shown that the total relativistic energy is given as Thus the relativistic energy can be computed in a more efficient method b Following the method outlined in part a, for a velocity of 09c, γ = 229 The relativistic momentum is therefore, c For a velocity of 095c, γ = 32 The relativistic momentum is therefore, d For a velocity of 099c, γ = 709 The relativistic momentum is therefore, e For a velocity of 0999c, γ = 224 The relativistic momentum is therefore,
3 From equation (2) we have the kinetic energy given by Using the fact that v << c, we can use the binomial expansion on the Lorentz factor for small arguments This gives Therefore the equation for the kinetic energy becomes which is the classical expression for kinetic energy 244 Will be posted once all of the homework has been turned in 247 For a photon, a its momentum is given by b its wavelength is given by the de Broglie relation c its frequency is given by 2413 The relativistic energy is given as The rest energy, mc 2 is 8199x10-19 J = 0511 MeV From the de Broglie wavelength we can calculate the momentum of the electron, Thus the relativistic energy is Wednesday, March 3, 2010 Chapter 24 Multiple-Choice
4 Problems 2411 A Compton Effect experiment a The wavelength momentum of the incident gamma ray are given as respectively b Using the Compton formula for the wavelength of the scattered photon the fact that the energy is inversely proportional to the wavelength we can write c The energy of the scattered gamma ray photon is mass of the electron is given as where the rest d The kinetic energy of the recoiling electron is found using conservation of energy where when the incident gamma ray photon interacts with electrons in the sample, the gamma ray photon loses some energy to the electron as it scatters Thus the kinetic energy of the recoiling electron is e The speed of the recoiling electron is given by using the expression for the relativistic kinetic energy We have 2412 X-rays on a foil target a The energy is given by b The scattered wavelength is given by
5 Thus, The energy is given by the formula in part a, is 118x10-14 J = MeV c The energy given to the foil is
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