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

Question

An α -particle after passing through potential difference of V volt collides with a nucleus. If the atomic number of the nucleus is Z , then distance of closest approach is, 14 . 4 Z V Å, 14 . 4 Z V m, 14 . 4 V Z m, 14 . 4 V Z Å

MARKS App · Accepted Answer

Given KE of α -particle = 2 eV r = Z e e 4 π ε 0 · KE = 2 Z e × 9 × 10 9 2 V ⇒ r = 2 × Z × 16 × 10 - 19 × 9 × 10 9 2 V = 14 . 4 · Z V Å

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