A proton is located at the origin, and a second proton is located on the x-axis

Question

A proton is located at the origin, and a second proton is located on the x-axis at x = 5.46 fm (1 fm = 10-15 m). (a) Calculate the electric potential energy associated with this configuration. J (b) An alpha particle (charge = 2e, mass = 6.64 10-27 kg) is now placed at (x, y) = (2.73, 2.73) fm. Calculate the electric potential energy associated with this configuration. J (c) Starting with the three particle system, find the change in electric potential energy if the alpha particle is allowed to escape to infinity while the two protons remain fixed in place. (Throughout, neglect any radiation effects.) J (d) Use conservation of energy to calculate the speed of the alpha particle at infinity. m/s (e) If the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity. m/s

Explanation / Answer

example Electrical potential energy (alpha particle)? A proton is located at the origin, and a second proton is located on the x-axis at x = 5.84 fm (1 fm = 10-15 m). (a) Calculate the electric potential energy associated with this configuration. .000000000000039461 J (I figured out (a) already using V=kq/r formula then I converted the V to J) (b) An alpha particle (charge = 2e, mass = 6.64 10-27 kg) is now placed at (x, y) = (2.92, 2.92) fm. Calculate the electric potential energy associated with this configuration. _____J I'm having trouble with this problem (b) (c) Starting with the three particle system, find the change in electric potential energy if the alpha particle is allowed to escape to infinity while the two protons remain fixed in place. (Throughout, neglect any radiation effects.) ________J (d) Use conservation of energy to calculate the speed of the alpha particle at infinity._______m/s (e) If the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity.______m/s ans The potential energy at the alpha particle is K*2*e²/(v[x² + y²] + K*2*e²/v[x² + y²] = K*4*e²/(x*v2), where x = y = 2.92*10^-15 m. The potential energy at the protons is K*e²/(5.54*10^-15) + K*2*e²/(x*v2). The potential energy of the alpha particle at 8 is 0, so the change = the potential energy in its initial position (calculated above). To find the speed, equate that to 0.5*m*v² and solve for v, where m = mass of the alpha particle The potential energy of the protons are equal and was calculated above; again equate this to 0.5*m*v² and solve for v, using m as the mass of the proton.

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