ch24_p51 - graph as indicated above). Thus, it must reach a...

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5 1. (a) Using U = qV we can “translate” the graph of voltage into a potential energy graph (in eV units). From the information in the problem, we can calculate its kinetic energy (which is its total energy at x = 0) in those units: K i = 284 eV. This is less than the “height” of the potential energy “barrier” (500 eV high once we’ve translated the
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Unformatted text preview: graph as indicated above). Thus, it must reach a turning point and then reverse its motion. (b) Its final velocity, then, is in the negative x direction with a magnitude equal to that of its initial velocity. That is, its speed (upon leaving this region) is 1.0 10 7 m/s....
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