Question

In a Geiger-Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when an α-particle of 8 MeV energy impinges on it before it comes to momentarily rest and reverses its direction. How will the distance of closest approach be affected when the kinetic energy of the α-particle is doubled?

Answer 1

r₀= closest approach to nucleus

Within this distance, K.E of the alpha particle is converted to potential energy.

Given, Z= 80, K.E= 8 MeV

K.E= (1/4πε₀){(Ze)(2e)/r₀}

   r0= (1/4πε₀){(Ze)(2e)/K.E}

      ={9×10⁹×2×80×(1.6×10⁻¹⁹)²}/ (8×106×1.6×10⁻¹⁹)

      = 2.88×10⁻¹⁴

As we know, r₀∝ (1/K.E)

So,  if the distance of the closest approach is halved, Kinetic energy becomes doubled.