An alpha particle ( two protons, two neutrons) moves into a stationary gold atom (79 protons, 118 neutrons), passing through the electron region that surrounds the gold nucleus like a shell and headed directly toward the nucleus (Fig. 24-42 . The alpha particle slows until it momentarily stops when its center is at radial distance `r=9.23` fm from the nuclear center. Then it moves back along its incoming path. (Because the gold nucleus is much more massive than the alpha particle, we can assume the gold nucleus does not move.) What was the kinetic energy `K_(i)` of the alpha particle when it was initially far away (hence external to the gold atom)?. Assume that the only force acting between the alpha particle and the gold nucleus is the (electrostatic) Coulomb force and treat each as a single charged particle.

Explanation

During the entire process, the mechanical energy of the alpha particle + gold atom system is conserved.
Reasoning: When the alpha particle is outside the atom, the system.s initial electric potential energy `U_(i)` is zero because the atom has an equal number of electrons and protons, which produce a net electric field of zero. However, once the alpha particle passes through the electron region surrounding the nucleus on its way to the nucleus, the electric field due to the electrons goes to zero. The reason is that the electrons act like a closed spherical shell of uniform negative charge and, as discussed in Section 23.4, such a shell produces zero electric field in the space it encloses. The alpha particle still experiences the electric field of the protons in the nucleus, which produces a repulsive force on the protons within the alpha particle.
As the incoming alpha particle is slowed by this repulsive force, its kinetic energy is transferred to electric potential energy of the system. The transfer is complete when the alpha particle momentarily stops and the kinetic energy is `K_(f) =0`.
Calculations: The principle of conservation of mechanical energy tells us that
`K_(i) + U_(i) = K_(f) + K_(r)" "`(24-63)
We know two values: `U_(i) =0 and K_(f) =0`. We also know that the potential energy `U_(f)` at the stopping point is given by the right side of Eq. `24-61`, with `q_1= 2e, q_2 =79e` (in which `e` is the elementary charge, `1.60 xx 10^(-19) C)`, and `r=9.23` fm. Thus, we can rewrite Eq. `24-63` as
`K_(i) = (1)/( 4 pi epsilon_(0) ) ((2 e)(79 e))/(9.23 fm)`
`=((8.99 xx 10^(9) N. m^(2) // C^(2) ) (158) (1.60 xx 10^(-19) C)^(2) )/(9.23 xx 10^(-15) m)`
`=3.94 xx 10^(-12) J =24.6 Me V." "` (Answer)