An α - particle after passing through a potential difference of V volts collides with a nucleus. If the atomic number of the nucleus is Z then the distance of closest approach of α - particle to the nucleus will be

Question

An α - particle after passing through a potential difference of V volts collides with a nucleus. If the atomic number of the nucleus is Z then the distance of closest approach of α - particle to the nucleus will be, 14.4 Z V A °, 14.4 Z V m, 14.4 Z V c m, all of these

MARKS App · Accepted Answer

K.E. = P.E. = qV ⇒ 2 e V = K Z e 2 e d = q V ∴ d = 9 × 1 0 9 × Z × e × 2 e 2 e V ∴ d = 9 × 1 0 9 × 1.6 × 1 0 – 19 × Z V d = 14.4 × 1 0 - 10 Z V m

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