Question

A proton is located at the origin, and a second proton is located on the x-axis at x₁ = 5.56 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 (x₂, y₂) = (2.78, 2.78) 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

Answer 1

a) U1 = k*q1*q2/r = (9e9*1.6e-19*1.6e-19)/(5.56e-15) = 4.14*10^-14 J

b)

r2 = r1 = sqrt ( (x2-x1)^2 + (y2-y1)^2 ) = 3.93 fm

U = k*q1*Q2/r + k*q1*q3/r1 + k*q2*q3/r1

U = (9e9*1.6e-19*1.6e-19)/5.56e-15 + (9e9*2*1.6e-19*1.6e-19)/3.93e-15) + (9e9*2*1.6e-19*1.6e-19)/3.93e-15)

U = 4.14e-15 + 1.17e-13 + 1.17e-15

U2 = 1.22e-13 J

c) dU = U2 - U1 = 1.1817e-13 J

d) dU = KE

1.1817e-13 = 0.5*6.67e-27*v^2

v = 5.95*10^6 m/s

e) U2 = KE

1.22e-13 = 0.5*m*v^2 + 0.5*m*v^2

1.22e-13 = m*v^2 = 1.67e-27*v^2

v = 8.54*10^6 m/s