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A photon decays into an electron-positron pair. What is the kinetic energy of the electron if its speed is 0.992 c size 12{c} {} ?

Answer

KE rel = ( γ 1 ) mc 2 = 1 1 v 2 c 2 1 mc 2 = 1 1 ( 0.992 c ) 2 c 2 1 ( 9.11 × 10 31 kg ) ( 3.00 × 10 8 m/s ) 2 = 5.67 × 10 13 J
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Section summary

  • Relativistic energy is conserved as long as we define it to include the possibility of mass changing to energy.
  • Total Energy is defined as: E = γmc 2 , where γ = 1 1 v 2 c 2 .
  • Rest energy is E 0 = mc 2 , meaning that mass is a form of energy. If energy is stored in an object, its mass increases. Mass can be destroyed to release energy.
  • We do not ordinarily notice the increase or decrease in mass of an object because the change in mass is so small for a large increase in energy.
  • The relativistic work-energy theorem is W net = E E 0 = γ mc 2 mc 2 = γ 1 mc 2 .
  • Relativistically, W net = KE rel , where KE rel is the relativistic kinetic energy.
  • Relativistic kinetic energy is KE rel = γ 1 mc 2 , where γ = 1 1 v 2 c 2 . At low velocities, relativistic kinetic energy reduces to classical kinetic energy.
  • No object with mass can attain the speed of light because an infinite amount of work and an infinite amount of energy input is required to accelerate a mass to the speed of light.
  • The equation E 2 = ( pc ) 2 + ( mc 2 ) 2 relates the relativistic total energy E and the relativistic momentum p . At extremely high velocities, the rest energy mc 2 becomes negligible, and E = pc .

Conceptual questions

How are the classical laws of conservation of energy and conservation of mass modified by modern relativity?

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What happens to the mass of water in a pot when it cools, assuming no molecules escape or are added? Is this observable in practice? Explain.

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Consider a thought experiment. You place an expanded balloon of air on weighing scales outside in the early morning. The balloon stays on the scales and you are able to measure changes in its mass. Does the mass of the balloon change as the day progresses? Discuss the difficulties in carrying out this experiment.

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The mass of the fuel in a nuclear reactor decreases by an observable amount as it puts out energy. Is the same true for the coal and oxygen combined in a conventional power plant? If so, is this observable in practice for the coal and oxygen? Explain.

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We know that the velocity of an object with mass has an upper limit of c size 12{c} {} . Is there an upper limit on its momentum? Its energy? Explain.

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Given the fact that light travels at c size 12{c} {} , can it have mass? Explain.

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If you use an Earth-based telescope to project a laser beam onto the Moon, you can move the spot across the Moon’s surface at a velocity greater than the speed of light. Does this violate modern relativity? (Note that light is being sent from the Earth to the Moon, not across the surface of the Moon.)

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Problems&Exercises

What is the rest energy of an electron, given its mass is 9 . 11 × 10 31 kg ? Give your answer in joules and MeV.

8.20 × 10 14 J

0.512 MeV

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Find the rest energy in joules and MeV of a proton, given its mass is 1 . 67 × 10 27 kg .

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If the rest energies of a proton and a neutron (the two constituents of nuclei) are 938.3 and 939.6 MeV respectively, what is the difference in their masses in kilograms?

2 . 3 × 10 30 kg size 12{2 "." 3 times "10" rSup { size 8{ - "30"} } `"kg"} {}

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Questions & Answers

how did you get 1640
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If auger is pair are the roots of equation x2+5x-3=0
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MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
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please why isn't that the 0is in ten thousand place
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please why is it that the 0is in the place of ten thousand
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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