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Questions  >>  CBSE XI  >>  Chemistry  >>  Structure of Atom
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A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate the distance of closest approach for head on collision.

$\begin{array}{1 1} (a)\;1.137 \times 10^{-13}\;m\\(b)\;2.137 \times 10^{-13}\;m\\ (c)\;3.137 \times 10^{-13}\;m\\ (d)\;4.137 \times 10^{-13}\;m\end{array} $

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