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In a Geiger Marsden experiment, calculate the distance of closest approach to the nucleus of Z 80, when and particle of 8 MeV energy impinges on it before it comes momentarily to rest and reverses its direction.How will the distance of closest approach be affected when the kinetic energy of the O-particle is doubled ? |
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Answer» Solution :As per question Z = 80, energy of alpha-particle`K = 8MeV = 8 xx 1.6 xx 10^(-13) J`. At distance of CLOSEST approach `(r_(0))` the original KINETIC energy of alpha-particle is completely converted into potential energy i.e., `K= (1)/(4pi in_(0)) = ((Ze)(2e))/(r_(0)) =(2Ze^(2))/(4 pi in_(0) r_(0))` ` rArr ""r_(0) = (2Ze^(2))/(4 pi in_(0) K)= ( 2 xx80 xx (1.6 xx 10^(-19))^(2) xx 9 xx 10^(9))/(8xx1.6 xx 10^(-13)) = 2.9 xx 10^(-14) m` If kinetic energyof thealpha - particles bedoubled ( i.e.,K = 2k) thenas peraboverelationdistanceof closet approch will become one half of its previousvalue(i.e.,`r_(0)= (1)/(2) r_(0)`) provided that `r_(0)`is still greater than the nuclear radius. |
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