10 A proton is located at the origin and a second proton is

#10

A proton is located at the origin, and a second proton is located on the x-axis at xx = 6.38 fm (1 fm = 10^-15 m). Calculate the electric potential energy associated with this configuration. An alpha particle (charge = 2e, mass = 6.64 Times 10^-27 kg) is now placed at (X_2, y_2) = (3.19, 3.19) fm. Calculate the electric potential energy associated with this configuration. 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.) Use conservation of energy to calculate the speed of the alpha particle at infinity. If the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity.

Solution

distance of separation between the protons

        d = 6.38fm = 6.38 e-15 m

charge on each proton = 1.6e-19 C

PE of two charge configuration separated by a distance r is given by

U = (q₁q₂/4₀r)

a) PE of the system = (1.6e-19)²/4*3.14*8.854e-12*6.38e-15

                             = 3.61e-14 J

b) when an alpha particle is placed at (3.19fm,3.19fm), we will have two more interactions to add up to the previous configurationthe alpha particle is equidistant from each of the Protons

d = sqrt(3.19² +3.19²) = 4.51fm

charge on alpha particle = 2*1.6e-19 C

PE of each alpha-proton pair =

     (1.6e-19)* (2*1.6e-19)/4*3.14*8.854e-12*4.51e-15

     = 1.02 e-13 J

Total PE of the configuration

          = 2*1.02 e-13 + 3.61e-14

            = 24.01 e-14 J

c) change in PE when the alpha particle is allowed to escape to infinity

           = 24.01 e-14 – 3.61e-14 J

            = 20.04e-14 J

d) KE of the alpha particle = mv²/2

when it is at infinity the change in PE is equal to its KE

mv²/2 = 20.4 e-14 J

speed of the alpha particle

v = sqrt(20.4 *2/6.64e-27)

    = 7.84e+13 m/s

e) When the 2 protons escape to infinity the PE of the system becomes 0 and is shared by the two protons as KE

mass of proton m_p = 1.66e-27 kg

m_p v² = 24.01e-14 J

speed of each proton when reached infinity

v = sqrt(24.01e-14/1.66e-27) = 1.20 e+7 m/s