(Answered) An alpha particle of energy 4 kev 1 ev 16 x 10-19 j is

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An Alpha particle of energy 4 keV (1 eV = 1.6 x 10^-19 J) is scattered by an aluminium atom through an angle of 90 degrees. Calculate the distance of closest approach to the nucleus. ( Atomic number of alpha particle = 2, atomic number Al = 13, e = 1.6 x 10^-19 C).

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