A proton is located at the origin, and a second proton is located on the x-axis
ID: 1615429 • Letter: A
Question
A proton is located at the origin, and a second proton is located on the x-axis at x_1 = 6.36 fm (1 fm = 10^-15 m). (a) Calculate the electric potential energy associated with this configuration. (b) An alpha particle (charge = 2e, mass = 6.64 times 10^-27 kg) is now placed at (x_2, y_2) (3.18, 3.18) fm. Calculate the electric potential energy associated with this configuration. (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. (e) If the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity.Explanation / Answer
a)
q = charge on each proton = 1.6 x 10-19 C
r = distance between protons = 6.36 fm = 6.36 x 10-15 m
electric potential energy is given as
U = k q2/r = (9 x 109) (1.6 x 10-19)2 /(6.36 x 10-15)
U = 3.62 x 10-14 J
b)
r1 = distance between the proton at origin and alpha particle = sqrt(3.182 + 3.182) = 4.5 fm = 4.5 x 10-15 m
r2 = distance between other proton and alpha particle = sqrt((3.18 - 6.36)2 + 3.182) = 4.5 fm = 4.5 x 10-15 m
Q = charge on alpha particle = 2 e = 2 x 1.6 x 10-19 C
Total electric potential energy
U = kq2/r + k Qq/r1 + k Qq/r2
U = (9 x 109) (1.6 x 10-19)2 /(6.36 x 10-15) + 2 (9 x 109) (1.6 x 10-19)(2 x 1.6 x 10-19 ) /(6.36 x 10-15) since r1 = r2
U = 1.81 x 10-13 J
c)
change in PE = 3.62 x 10-14 - 1.81 x 10-13 = - 1.45 x 10-13 J
d)
kinetic energy at infinity = - change in PE = - ( - 1.45 x 10-13) = 1.45 x 10-13
(0.5) m v2 = 1.45 x 10-13
(0.5) (6.64 x 10-27) v2 = 1.45 x 10-13
v = 6.61 x 106 m/s
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