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WITH WHAT VELOCITY SHOULD AN ALPHA PARTICLE BE PROJECTED TOWARDS THE NUCLEUS OF COPPER ATOM SO THAT DISTANCE OF CLOSEST APPROACH IS 10^-13 m ?

WITH WHAT VELOCITY SHOULD AN ALPHA PARTICLE BE PROJECTED TOWARDS THE NUCLEUS OF COPPER ATOM SO THAT DISTANCE OF CLOSEST APPROACH IS 10^-13 m ?

Grade:11

1 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
9 years ago
Consider an α-particle - the nucleus of a He4 atom - travelling with velocity v and kinetic energy KE along the line connecting its centre with that of a stationary copper atom. As the two positively charged nuclei approach, the electrostatic repulsion force between them slows down the α-particle. The copper nucleus, being massive in comparison, may be assumed to remain stationary, as a first approximation. At some point the kinetic energy will fall to zero, and the total energy of the system will be the potential energy PE of the two charged particles a distance r apart. The nuclei then separate and, as the collision is elastic, retrace their previous trajectories.

From the principle of conservation of energy, KE = PE, and simple expressions for both are available from classical physics.

KE = ½.mα.v²

where mα is the mass of the α-particle, 6.644 x 10^(-27) kg, and
v is its initial velocity, 2.00 x 10^7 m/s.

PE = ke.[q1.q2/r]

where ke is the Coulomb constant, and ke = 1/[4.π.ε] where ε is the
permittivity of a vacuum - a defined value, 8.854 x 10^(-12) F/m.
The charges q1 and q2 of the α-particle and the copper nucleus respectively,

velocity=sqrt(9*10^9*2*29*2.56*10^-38*2/10^-13*4*6.6*10^-27)
=3.16*10^6m/s

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