Question

a) Calculate the electric potential energy associated with this configuration. (in Joules)

b) An alpha particle (charge=2e, mass=6.64 x 10^-27kg) is now placed at (x,y)=(3.32,3.32) fm.Calculate the electric potential energy associatedwith this configuration. (in Joules)

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 twoprotons remain fixed in place.(Throughout, neglect any radiation effects.) (in Joules)

d) Use conservation of energy to calculate the speed of the alpha particle at infinity. (in 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. (in m/s)

Answer 1

Given

distance between two protons d = 6.64 fm = 6.64*10¹⁵ m

Electrical potential energy

U = k q1 q2 / r

= ( 9*10⁹ )(1.6*10^-19 C)² / ( 6.65*10¹⁵ m )

=3.44*10^-44 J

The mutual electric potential of system of charges is

=(9*10⁹)[(1.6*10^-19)²/6.64+2(1.6*10^-19)²/4.695+2(1.6*10^-19)²/4.695]

= 2.309*10^-28 J

change in electric potential energy if the alpha particle is allowed to escape to infinity

=(9*10⁹)[2(1.6*10^-19)²/4.695+2(1.6*10^-19)²/4.695 ]

= 1.9629*10²⁸ J

So conservation of energy gives

K_f + U_f = K_i + U _i

( 1/2) m v ² = 1.9629*10²⁸J

where m is maass of alpha particle m = 6.64465620(33)×10−27
kg

plug the values and get the speed of alpha particle