Question

An α-particle of energy 5 MeV is scattered though 180° by a fixed uranium nucleus. The distance of the closet approach is of the order of 

(A) 10^–8 cm 

(B) 10^–12 cm 

(C) 10^–10 cm 

(D) 10^–15 cm

Answer 1

The correct option is (B) 10^–12 cm.

Explanation:

Given uranium nucleus ∴ Z = 92,

E_k = 5 MeV = 5 × 10⁶ × 1.6 × 10^–19 J

The distance of closet approach is r = [(k ∙ 2Ze²) / E_k]

∴ r = [{9 × 10⁹ × 2 × 92 × (1.6 × 10^–19)²} / (5 × 10⁺⁶ × 1.6 × 10^–19)]

= 5.29 × 10^–14­­ m

= 5.29 × 10^–12 cm

∴ order is 10^–12 cm