Calculate the nearest distance of approach of an `alpha`-particle of energy 2.5 Me V being scattered by a gold nucleus (Z=79).

Large angle scattering of alpha- particle led Rutherford to the discovery of atomic nucleus. It is the tiny central core of every atom in which entire positive charge of the atom is concentrated. The distance of closet approach of alpha particle from the nucleus is r_(theta)=((Ze)(2e))/(4pi epsilon_(theta)((1)/(2)mv^(2))) This distance r_(theta) gave him the order of size of nucleus. Read the above passage and answer the followign question: (i) What is the distance of closet approach of an alpha particle of energy 7.7 MeV from gold nucleus (z=79)? (ii) What is the implication of this relation in day to day life?

Calculate the distance of closest approach when a proton of energy 3 MeV approaches a gold nucleus (Z=79).

What is the impact parameter of an alpha -particle of initial energy 5 MeV scattered by 60^@ for Z=79 ?

What is the distance of closest approach to the nucleus for an alpha -particle of energy 5J MeV which undergoes scattering in the Gieger-Marsden experiment.

Calculate the distance of closest approach for a proton with energy 3 MeV approaching a gold nucleus.

In a Geiger - Marsden experiment, calculate the distance of closest approach to the nucleus of Z=75, when an alpha - particle of 5 Me V enery 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 alpha -particle is doubled ?

A closest distance of approach of an "alpha" particle travelling with a velocity "v towards "Al(13)" nucleus is d The closest distance of approach of an alpha particle travelling with velocity 4V towards Fe(26) nucleus is