Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
17
Electric Potential
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17-1 to 17-4: Electric Potential
17-5: Potential Due to Point Charges
17-6: Electric Dipoles
17-7: Capacitance
17-8: Dielectrics
17-9: Electric Energy Storage
17-10: Digital
17-11: TV and Computer Monitors

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 12
Q

What is the speed of an electron with kinetic energy

  1. 850 eV, and
  2. 0.50 keV?
A
  1. 1.7×107 m/s1.7 \times 10^7 \textrm{ m/s}
  2. 1.3×107 m/s1.3 \times 10^7 \textrm{ m/s}
Giancoli 7th Edition, Chapter 17, Problem 12 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Kinetic energy is one-half mass times velocity squared. So if you multiply both sides by two and divide both sides by m, you get two times kinetic energy over m is v squared. Then when you take the square root of both sides, you get v is the square root of two times kinetic energy divided by mass. So we have square root of two times 850 electron volts, which we convert into Joules by multiplying by 1.6 times 10 to the minus 19 Joules per electron volt, then divided by the rest mass of an electron, 9.11 times 10 to the negative 31 kilograms, this gives about 1.7 times 10 to the seven meters per second. If the electron had 0.5 kilo electron volts-- Careful with the tricky units that put kilo electron volts. That's 0.5 times 10 to the three electron volts multiplied by the same conversion from electron volts into Joules divided by the same mass of electron. Then you get 1.3 times 10 to the seven meters per second.

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