Science, asked by ashishraj95702e4276, 11 months ago

The distance of closest approach (in fm) of an a-particle travelling towards gold nucleus
(At. No. = 79) with kinetic energy 2.4 MeV.

Answers

Answered by psam26

2

Answer:

equate KE with PE..

use formula

r = Ze^2/ 2π€K

€= epsilon not

K= Kinetic energy

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