InterviewSolution
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InterviewSolution
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A proton is first from very loward a nucleus with charge `Q = 120 e , ` where e is the nucles The de Brogle wavelength (in unit of fin) of the proton at its start is (tke the proton mass , `m _(p) = (5//3) xx 10^(-27) kg h//s = 4.2 xx 10^(-15) J s // C,` `(1)/(4 pi s_(0)) = 9 xx 10^(9) m//F , 1 fm = 10^(-15) m ` |
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Answer» `1` cos in K.E. of proton = gain in potential energy of the proton - nucleus system `(1)/(2) m nu^(2) = (1)/(4 pi s_(0)) (q_(1) q_(2))/(r )` `:. (p^(2))/(2m) = (1)/(4 pi epsilon_(0))(q_(1) q_(2))/(r )` :. (1)/(2m)((h^(2))/(h^(2))) = (1)/(4 pi epsilon_(0))(q_(1) q_(2))/(r )` `:. lambda = sqrt((4 pi epsilon_(0) r , h^(2))/(q_(1) q_(2) (2m))) = 7 fm` |
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