Question

In a Geiger-Marsden experiment, calculate the distance of closest approach to the nucleus of $Z = 80$, when $\alpha$ -particle of $8\,MeV$ energy impinges on it before it comes momentarily to 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

In order to solve this question, we are going to compute the expression for the kinetic energy of the $\alpha$-particle, and the relation of distance of closest approach of the coleus and alpha particle is found, after that the impact of making the kinetic energy double on the distance of closest approach is calculated.

Complete step by step answer:
The atomic number of the nucleus is given: $Z = 80$. The energy of the$\alpha$-particle is equal to $8\,MeV$. Now if we consider the distance of closest approach between the nucleus and the $\alpha$-particle equal to${r_0}$. The kinetic energy of the alpha particle is given as :
${E_k} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{\left( {Ze} \right)\left( {2e} \right)}}{{{r_0}}}$
The inference that can be drawn from the above relation is that the distance of the closest approach is inversely proportional to the kinetic energy of the$\alpha$-particle:
${r_0} \propto \dfrac{1}{{{E_k}}}$
Hence, we can conclude that when the kinetic energy of the $\alpha$-particle, the distance of closest approach between the nucleus and the $\alpha$-particle is halved.

Note: The distance of closest approach of the nucleus and the$\alpha$-particle is the distance between their centers when they are externally tangent. In the Geiger Marsden-experiment, Rutherford sent a beam of alpha particles (helium nuclei) emitted from a radioactive source against a thin gold foil.