Question

help with this question part A-E this is all ONE QUESTION. all parts 3

A proton is located at the origin, and a second proton is located on the x-axds at x w 6.34 fm (1 fm 1015 m). 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 neglect any radiation effects.) m/s m/s O Type here to search

Answer 1

PE_a = k q_1 q_2/r_12 = (9 times 10^9) (1.6 times 10^-19)^2 = 3.634 times 10^-14 + 2 times (9 times 10^9) (1.6 times 10^-19) (2 times 1.6 times 10^-19)/squareroot (3.17)^2 + (3.17)^2 times 10^-15 = 2.419 times 10^-13 J Delta P E = 3.634 times 10^-14 - 2.419 times 10^-13 = -2.0556 times 10^-13 j By conservation of vector 1/2 mv^2 = -Delta P R v = squareroot 2 times 2.0556 times 10^-13/6.64 times 10^-4 v = 7.8686 times 10^6 m/s

e)

By conservation of energy

(1/2)m_pV² =(1/2)PE

V=sqrt[2.419*10^-13/1.67*10^-27]

V=1.204*10⁷ m/s