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Numerade Educator



Problem 50 Hard Difficulty

(a) Show that a potential difference of $1.02 \times 10^{6} \mathrm{V}$ would be sufficient to give an electron a speed equal to twice the speed of light if Newtonian mechanics remained valid at high speeds. (b) What speed would an electron actually acquire in falling through a potential difference equal to $1.02 \times 10^{6} \mathrm{V} ?$


a) 2$c$
b) 0.943$c$


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Video Transcript

we know it. Speed. We is equal to square root off. Uh, two times kinetic energy divided by Mm. On dhe this came direct unease, uh, squared route off to into since kinetic energies charge multiply by potential difference. Delta bi divided by, um, and putting the values we have, uh, sweet loot. D'oh Into kill. Well, cumin is 1.6 multiplied by 10 to the dollar minus 19. Multiply by 1.21 point 02 Multiply by 10 to the power 10 to the dollar six divided by 9.11 multiplied by 10 to the power minus 31. And solving these we is ah, 5.99 My find mine mine most of line by tend to the power eight, which is twice the speed of light. All right, also, uh, you know, they kinetic energy is equal to gum. A minus one times. Uh, e r e r is rest Marcin Chief, that in this case, kinetic energies 1.2 million electron won't is equal to gonna minus one. Multiplied by zito. Find 511 million electron volts and solving for gamma gamma is equal to three governments the relativistic factor, but he is equal to Ah, square root off, uh, one minus one, divided by, uh, Kama Square times uh, the speed of light. See, and putting the values we is equal to the square root off Well, uh, spirit Rudolph one minus one, divided by three whole squared, multiplied by C and therefore, this is equal to zero point line for three times the speed of light.

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