An `alpha`-particle having kinetic energy 5 MeV falls on a Cu-foil. The shortest distance from the nucleus of Cu to which `alpha`- particle reaches is (Atomic no. of Cu=29,`K=9xx10^(9)Nm^(2)//C^(2)`)

An alpha -particle having kinetic energy x MeV falls on a Cu-foil. The shortest distance from the nucleus of Cu to which alpha -particle reaches is 1.67 xx 10^(-14) m . What is the value of x ? [Given : Atomic number of Cu = 29, K = 9 xx 10^(9) Nm^(2)//C^(2) ]

An alpha -particle having energy = 10 meV collides with a nucleus of atomic number 50. The distance of the closest approach is

An alpha particle having kinetic energy equal to 8.7 MeV is projected towards the nucleus of copper with Z=29.Calculate its distance of closet 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 2 xx 10^(6) V falls on a silver foil. The atomic number of silver is 47. Calculaye (i) the kinetic energy of the alpha -particle at the time of falling on the foil (ii) the kinetic energy of the alpha -particle at a distance of 5 xx 10^(-14) m from the nucleus and (iii) the shortest distance from the nucleus of silver to which the alpha -particle reaches.

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 of energy 5 MeV is moving forward for a head on collision. The distance of closest approach from the nucleus of atomic numbe Z=50 is ... xx10^(-14)m (k=9xx10^(9)SI, c=1.6xx10^(-19)C) ,