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What percentage of the speed of light will an electron be moving after being accelerated through a potential difference of 2,295 Volts?

User Leonidas
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1 Answer

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In order to find the answer, we must relate electrical potential energy with kinetic energy

1/2(m)(v^2) = eV

v = sqrt(2eV/m) = 28408352 m/s

speed of light = 299 792 458 m / s

28408352/299 792 458 = 0.09475 or 9.475 %

User Sty
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