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In a Geiger-Marsden experiment, calculate the distance of the closest approach to the nucleus of Z = 80, when an a-particle of 8 MeV energy impinges on it before it comes momentarily to rest and reverses its direction. How will the distance of the closest opproach be affected when the kinetic energy of the o-particle is doubled? |
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Answer» Solution :`((Ze)(2e))/(3pi epsi_(0)(r_(0)))=E` `r_(0)=(2Ze^(2))/(4pi epis_(0)(E))=(2Ze^(2))/(4pi epsi_(0)xxE)` `r_(0)=(9xx10^(9)xx2xx80xx(1.6xx10^(-19))^(2))/(8xx10^(6)xx(1.6xx10^(-19)))m` `=(18xx1.6xx10^(-10)xx80)/(8xx10^(6))` `=2.88xx10^(-14)m` `r_(0) prop (1)/(KE)` If K.E.becomes twice then `r_(0)=(r_(0))/(2)` i.e., distance of closet apporach becomes HALF. |
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