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An alpha -particle moving with initial kinetic energy 'K' towards a nucleus of atomic number Z approaches a distance 'r_(0)' at which it reverses its direction. Obtain the expression for the distance of closest approach 'r_(0)' in terms of kinetic energy of a particle. |
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Answer» Solution :Consider an o-particle of mass m, charge + 2e and an initial speed v (or initial KINETIC energy K) moving straight along the central LINE of a nucleus of atomic number Z. As `alpha`particle approaches the nucleus, on account of electrostatic repulsion, its speed goes on DECREASING and at a point SITUATED at a distance `r_(0)`, the `alpha`-particle momentarily comes to REST. Obviously in this position the initial kinetic energy of `alpha`-particle has been completely converted into electrostatic potential energy i.e., `K.E ` at`A =P.E` at B `therefore "" k = (1)/(2) mv^(2)= (1)/(4 pi in_(0)) .((+2e)(+Ze))/(r_(0))` `rArr "" r_(0) = (1)/(4 pi in_(0)) = (2Ze^(2))/(((1)/(2) mv^(2))) =(1)/(4pi in_(0)) .(2Ze^(2))/(K)` 
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