Question

An α particle fired head-on at a stationary nickel nucleus approaches to a radius of 24 fm before being turned around. What is the maximum Coulomb force exerted on the α particle? What is the electric potential energy of the α particle at its point of closest approach? Find the initial kinetic energy of the α particle.

Answer 1

F_coulomb = kq₁q₂/r²

q₁=charge on alpha particle = 2e

q₂=charge on nickel nucleus = 28e

r= distance between q₁ & q₂

F_coulomb = (8.99*10^9*2*1.6*10^-19*28*1.6*10^-19)/(24*10^-15)²

F_coulomb = 22.37 N

U_f = kq₁q₂/r

U_f = (8.99*10^9*2*1.6*10^-19*28*1.6*10^-19)/(24*10^-15)

U_f = 5.37*10^-13 J

ΔKE = - ΔU

KE_f – KE_i = -(U_f – U_i)

KE_f – KE_i = U_i – U_f

At closest approaches distance v_f=0m/s => KE_f = 0 J

At initial point r=infinity => U_i = 0 J

Thus,

0– KE_i = 0– U_f

KE_i = U_f

Plug above values of U_f

KE_i = 5.37*10^-13 J