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The distance of the closest approach of an alpha- particle fired at a nucleus with kinetic energy K is r_(0). The distance of the closest approach when the alpha-particle is fire at the same nucleus with kinetic energy 2K will be |
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Answer» `(r_(0))/(2)` `r_(0)=(2Ze^(2))/(4pi epsi_(0)xxK)` where `2kze^(2)` are constant `:.r_(0)= prop (1)/(K)` `:. ((r_(0))_(2))/((r_(0))_(1))=(K_(1))/(K_(2))=(K)/(2K)` `:.(r_(0))_(2)=((r_(0))_(1))/(2)=(r_(0))/(2) [ :. (r_(0))_(1)=r_(0)]` |
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