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Introductory Physics Homework Help
Finding the kinetic energy of a proton in megaelectron volts?
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[QUOTE="Violagirl, post: 4535931, member: 437260"] [h2]Homework Statement [/h2] Charged particles such as protons are accelerated to high velocities and are allowed to collide with atomic nuclei to probe their internal structure. The electric potential outside of a point charge Ze. A) A lead nucleus (Z - 82) is approximately described as a sphere of radius 7 x 10[SUP]-15[/SUP] m. How much kinetic energy in megaelectron volts (1Me V = 10[SUP]6[/SUP] electron volts) must a proton have to overcome the electrical repulsion and reach its surface? B) What is the corresponding initial velocity? [h2]Homework Equations[/h2] Kinetic energy: 1/2mv[SUP]2[/SUP] = e Δ V V = U/q U = W = F * D F = q * E E = kq/r[SUP]2[/SUP] [h2]The Attempt at a Solution[/h2] I posted all key equations above that I used to try to solve for part a. So I started out by solving for E: E = kQ/r[SUP]2[/SUP] = (9.0 x 10[SUP]9[/SUP] N m[SUP]2[/SUP]/r[SUP]2[/SUP] (1.6 x 10[SUP]-19[/SUP] C)/(7 x 10[SUP]-15[/SUP] m)[SUP]2[/SUP] = 2.93 x 10[SUP]19[/SUP] N/C From there, I used E to find F: F = q * E = (1.6 x 10[SUP]-19[/SUP] C) (2.93 x 10[SUP]19[/SUP] N/C) = 4.688 N From there, I found U: U = (F) ( D) = (4.688 N) (7 x 10[SUP]-15[/SUP] m) = 3.2816 x 10[SUP]-14[/SUP] J Finally, solved for V: V = U/q = (3.2861 x 10[SUP]-14[/SUP] J)/(1.6 x 10[SUP]-19[/SUP] C = 2.05 x 10[SUP]5[/SUP] V Solving for megavolts: 2.05 x 10[SUP]5[/SUP] (1 meV/10[SUP]6[/SUP] V) = .205 MeV However, this turned out to be wrong. The answer in my solutions manual says that it should be 16.8 MeV. What did I do wrong in this problem? Otherwise I figured for part b, I would use the equation: 1/2mv[SUP]2[/SUP] = eV to find v, the initial velocity. But can't quite yet without MeV for part a... [/QUOTE]
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Finding the kinetic energy of a proton in megaelectron volts?
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