Suppose that an alpha particle of 3.20 MeV approaches head-on a lead nucleus (Z = 82). Assuming that the lead nucleus remains at rest and the alpha particle momentarily comes to rest and reverses its direction at a distance much more than the radius of the lead nucleus, the distance of its closest approach is:

Concept:

• Alpha Particle:​​
  • ​Alpha particles are alternatively known as Alpha radiation or Alpha rays.
  • It is a positively charged particle emitted from the decay of various radioactive materials.
  • Alpha particle mass is due to the two protons and two neutrons bonding.
  • Thus, the Alpha ray nucleus is very similar to the Helium-4 nucleus.
• ​Distance of closest approach:
  • ​It is defined as the minimum distance of the charged particle at which the initial kinetic energy of the particle is equal to potential energy due to the charged nucleus. 
• The formula for the distance of the closest approach, \(d =\frac{1}{4\pi \epsilon_0}\frac{Ze^2}{K}\) 
• where K = kinetic energy, Z = atomic mass, e = electronic charge
• 1fm = 1.0 × 10^-15 m

Calculation:

Given,

The kinetic energy of the alpha particle, K = 3.20 MeV, The atomic mass, Z = 82

Let's consider r is the center-to-center between the alpha particle and nucleus. When the alpha particle is at the stopping point, then

\(K=\frac{1}{4\pi \epsilon_0 } \frac{(Ze)(2e) }{d}\)

\(d=\frac{1}{4\pi \epsilon_0 } \frac{2Ze^2 }{K}\)

\(d=9× 10^9× \frac{2× 82× (1.6× 10^{-19})^2 }{3.20× 10^{6}× 1.6× 10^{-19}}\)

d = 73.8 fm

Hence, the distance of its closest approach is 73.8 fm.