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An alpha particle of 10 MeV is moving forward fo a head on collision. What will be the distance of closest approach from the nucleus o atomic number Z = 29 ? |
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Answer» `8.4xx10^(-15)cm` Potential energy of `alpha`-particle at distance d=kinetic energy of `alpha`- particle at large distance `10xx10^(6)xx1.6xx10^(-19)=(K(2qe)(2e))/(r_(0))` `:.r_(0)=(9xx10^(9)xx29xx2xx(1.6xx10^(-19))^(2))/(16xx10^(-13))` `:.r_(0)=83.52xx10^(-16)` `:.r_(0)=8.4xx10^(-15)m` |
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