Question

An alpha-particle having energy 10Mev
collides with a nucleus of
Atomic number 50 . Then distance of
closest approach will be
$a.14.4 \times {10}^{- 16} m$
$b.1.7 \times {10}^{ - 7} m$
$c.1.5 \times {10}^{ - 12} m$
$d.14.4 \times {10}^{ - 15m}$


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Answer 1

Explanation:

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