Question

Determine the distance of closest approach when an alpha particle of kinetic energy 4.5 MeV strikes a nucleus of Z = 80, stops and reverses its direction.

Answer 1

Here, 

k.E of alpha particle, 

E = 4.5 MeV

= 4.5 × 1.6 × 10⁻¹³J 

= 7.2 × 10⁻¹³J

Z = 80, 

e = 1.6 × 10⁻¹⁹C

∴ Distance of closest approach,