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In a Geiger - Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when an α particle of 8 MeV energy impinges on it before it comes momentarily to rest and reverses its direction.

How will the distance of closest approach be affected when the kinetic energy of the α is doubled?

(ze)(2e)4πε0(r0)=kr0=2Ze24πε0K                                                                  (1) marks

r0=9×109×2×80×(1.6×10-19)28×106×(1.6×10-19)m                              (1/2) marks

  =18×1.6×10-10×808×106                                                (1/2) marks

 =2.88×10-14 m                                                            (1/2) marks

r01KE

If KE becomes twice then r0'=r02

i.e. distance of closest approach becomes half                   (1/2) marks

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