Question

-/5 points SerCP1016.P.019. My Notes Ask Your A proton is located at the origin, and a second proton is located on the x-axis at, -5.84 m (1 fm (a) Calculate the electric potential energy associated with this configuration 10"" m). (b) An alpha particle (charge - 2e, mass - 6.64 x 10"" kg) is now placed at (x,y) - (2.92, 2.92) 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 my'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
help

Answer 1

1.

(a)

electric potential energy = k*q1*q2/r
q1 = q2 = 1.6* 10^-19
k = 9 *10⁹
r = 5.84 * 10^-15

potential energy , V₁₂= 9*10⁹*(1.6*10^-19)²/5.84*10^-15

potential energy V₁₂ = 3.94*10^-14 J
b)
alpha particle charge = 3.2*10^-19
total potential energy U = V₁₂ + V₂₃ +V₃₁

r₂₃ = r₃₁ = 4.12*10^-15 m
V₂₃ = V₃₁ = 9*10⁹*2*1.6*10^-19*1.6*10^-19/(4.12*10^-15) = 11.18*10^-14 J
total potential energy = 2*11.18*10^-14 + 3.94*10^-14

total potential energy, U = 26.30*10^-14 J

c)

dU = U - V₁₂

dU = 22.36*10^-14 J

d)

dU = dE

22.36*10^-14 = 0.5*6.64*10^-27*v²

v = 2.59*10⁷ m/s

e)

U = KE

26.30*10^-14 = 0.5*m*v² + 0.5*m*v²

26.30*10^-14 = m*v²

v = 1.99*10⁷ m/s