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An alpha- particle of 10 MeV is moving forward for a head on collision. What will be the distance of closest approach from the nucleus of atomic number Z = 50 ? |
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Answer» `1.44xx10^(-14)m` Potential energy of a-PARTICLE at DISTANCE d = kinetic energy of a particle at LARGE distance `(1)/(4piepsi_(0))((Ze)(2e))/(d)=10MeV` `d=((Ze)(2e))/(4piepsi_(0)xx10xx10^(6)xx1.6xx10^(-19))` `=(2xx50xx(1.6xx10^(-19))^(2))/(4xx3.14xx8.85xx10^(-12)xx10^(7)xx1.6xx10^(-19))` `=(100xx1.6xx10^(-19))/(4xx3.14xx8.85xx10^(-5))` `:.d=1.44xx10^(-14)m` |
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