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Wave Optics

Question
CBSEENPH12039819

When an alpha particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on mass as,

  • fraction numerator 1 over denominator square root of straight m end fraction
  • 1 over straight m squared
  • m

  • 1 over straight m

Solution

D.

1 over straight m

When an alpha particle moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy.
Initial Kinetic energy of the alpha particle = Potential energy of alpha particle at closest approach.
That is,
rightwards double arrow space 1 half m v squared space equals space fraction numerator 2 Z e squared over denominator 4 pi epsilon subscript o r subscript o end fraction

rightwards double arrow space r subscript o space proportional to space 1 over m
This is the required distance of closest approach to alpha particle from the nucleus.

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