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A proton is located at the origin, and a second proton is located on the x-axis at x,-6.50 fm (1 fm-10-15 m) (a) Calculate the electric potential energy associated with this configuration. 10-27 kg b) An alpha particle charge configuration. 2e, mass = 6.64 is now placed at x2 y2 325 3.25 m Calculate he electric potentia energy associated with this (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.) (d) Use conservation of energy to calculate the speed of the alpha particle at infinity. ารู้ไว้ (e) If the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity. m/sI know A is 3.544x10^-14 but I do not know how to solve for the rest

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The electrostatic potential energy is (9x109)(1.6x10-19)2 650× 10-15 = 3.544x10-14 J The distance is 3.25)(3.25) -4.596 fm 4.60× 10-15 m The potential energy is (( 98 10% )(1.6×10-19 )2(1.6×10-19 4.596x10-15 ) 3.544x1014+2 - 2.36x10-3 j The energy in electrostatic potential energy U2-U,-2.36x101-3.544x101 2.0x1013 J Then U-U1-^mvi 2.0 × 10-13--( 6.64 × 10-27 ) V2 v= 7.76×106 m/s Then change in electrostatic potential energy is 2.36x10136.64x102v v= 5.96× 106 m/s

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