A proton is located at the origin, and a second proton is located on the x-axis
ID: 1593116 • Letter: A
Question
A proton is located at the origin, and a second proton is located on the x-axis at x1 = 5.60 fm (1 fm = 1015 m).
(a) Calculate the electric potential energy associated with this configuration. J
(b) An alpha particle (charge = 2e, mass = 6.64 1027 kg) is now placed at (x2, y2) = (2.80, 2.80) 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
(a) Electrostatic potential energy is given by the following formula:
V = k*q1*q2/r
Here k= 9*10^9, q1 = q2 = 1.6*10^-19 = charge of proton, r = distance between the protons = 5.6 fm
Plugging in all values in to equation, we get,
V = 9*10^9*(1.6*10^-19)^2/(5.6*10^-15) = 4.11*10^-14 J
(b) Now for three particles, total electrostatic potential energy will be:-
V = V12+V23+V31 where,
V12 = P.E due to (q1 and q2), V13 = (q1 and alpha particle) and V23 = (q2 and alpha particle)
V12 = 4.11*10^-14 J
V23 = V13 = kq1q3/d3 = 9*10^9*2*(1.6*10^-19)^2/(sqrt(2.8^2+2.8^2)*10^-15) = 1.16*10^-13 J
V = 4.11*10^-14 + 2*1.16*10^-13 = 2.73*10^-13 J
(c) If alpha particle is allowed to escape to infinity it will not contribute anything to the total potential energy so the change in potential will be the difference between result of (a) and (b).
delta_V = 2.73*10^-13 - 4.11*10^-14 = 2.32*10^-13 J
(d) delta_V = 0.5*m_alpha_v^2
v = sqrt(2*V_delta/m_alpha) = sqrt(2*2.32*10^-13/(6.64*10^27)) = 8.36*10^6 m/s
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