Question 1.10.9: Consider a head-on collision between an α-particle and a lea...

In this head-on collision the distance of closest approach r_0  can be obtained from the conservation of energy E_i=E_f,  where E_i  is the initial energy of the system, α-particle plus the lead nucleus, when the particle and the nucleus are far from each other and thus feel no electrostatic potential between them. Assuming the lead nucleus to be at rest, E_i  is simply the energy of the α-particle: E_i=9.0MeV=9\times 10^{6}\times 1.6\times 10^{-19}J.
As for E_f,  it represents the energy of the system when the α-particle is at its closest distance from the nucleus. At this position, the α-particle is at rest and hence has no kinetic energy. The only energy the system has is the electrostatic potential energy between the α-particle and the lead nucleus, which has a positive charge of 82e. Neglecting the recoil of the lead nucleus and since the charge of the α-particle is positive and equal to 2e, we have E_f=(2e)(82e)/(4\pi \varepsilon _0r_0).  The energy conservation E_i=E_f  or  (2e)(82e)/(4\pi \varepsilon _0r_0)=E_i  leads at once to

r_0=\frac{(2e)(82e)}{4\pi \varepsilon _0E_i}=2.62\times 10^{-14}m,               (1.192)

where we used the values e=1.6\times 10^{-19}C  and  1/(4\pi \varepsilon _0)=8.9\times 10^{9}  N  m^2C^{-2}.