An alpha - particle of 5MeV energy strikes with a nucleus of uranium at stationary at an scattering angle of 180o .The nearest distance up to which alpha - particle reaches the nucleus will be of the order of

An alpha -particle of 5MeV energy strikes with a nucleus of uranium at stationary at an scattering angle of 108^(@) . The nearest distance up to which alpha particle reaches the nuvles will be of the order of

What is the impact parameter at which the scattering angle is 10^(@) for Z = 79 and initial energy of the alpha -particle is 5MeV ?

An alpha -particle of kinetic energy 7.68 MeV is projected towards the nucleus of copper (Z=29). Calculate its distance of nearest approach.

An alpha -particle with kinetic energy K is heading towards a stationary nucleus of atomic number Z . Find the distance of the closest approach.

An alpha - particle after passing through a potential difference of V volts collides with a nucleus . If the atomic number of the nucleus is Z then the distance of closest approach of alpha – particle to the nucleus will be

An alpha - particle after passing through a potential difference of V volts collides with a nucleus . If the atomic number of the nucleus is Z then the distance of closest approach of alpha – particle to the nucleus will be

alpha -particles of enegry 400 KeV are boumbardel on nucleus of ._(82)Pb . In scattering of alpha -particles, it minimum distance from nucleus will be

An alpha -particle moving with a constant energy is scattered by the nuclues. The scattering angle will be maximum when the alpha - particles.