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Find the distance of closest approach of an 5.0-MeV alpha particle incident on a gold foil.

Find the distance of closest approach of an 5.0-MeV alpha particle incident on a gold foil.

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Find the distance of closest approach of an 5.0-MeV alpha particle incident on a gold foil.
Transcribed Image Text:Find the distance of closest approach of an 5.0-MeV alpha particle incident on a gold foil.
Expert Solution
Concept and Principle:
  • Since the nucleus is positively charged any positively charged particle getting close will experience a repulsive force. This will reduce the kinetic energy of the particle. 

 

  • The point at which the kinetic energy of the particle becomes zero and its potential energy becomes maximum is the closest point the charged particle can reach.

 

  • Thus we have the distance of the closest approach as,

d=14πε0zZe2K

Here ε0 is the permittivity of free space, z is the atomic number of the projectile, Z is the atomic number of the target, e is the charge of the electron, and K is the kinetic energy of the projectile.

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