An $α$ -particle of energy 5 MeV is scattered through $180^{\circ}$ by gold nucleus. The instance of closest approach is of the order of

$10^{- 14} c m$

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$10^{- 10} c m$

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$10^{- 16} c m$

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$10^{- 12} c m$

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Solution

The correct option is D
$10^{- 12} c m$

Distance of closest approach is = $10^{- 14} m = 10^{- 12} c m$ (Order of size of nucleus)

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