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

Step-by-step solution

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