Question

An α -particle of energy 5 Me V is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of ______.

विकल्प

• 10^-12 cm

• 10^-10 cm

• 10^-20 cm

• 10^-15 cm

Answer 1

An α -particle of energy 5 Me V is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of 10^-12 cm.

Explanation:

Distance of closest approach

r₀ = `("Ze"(2"e"))/(4piepsilon_0(1/2"mv"^2))`

Energy, E = 5 × 10⁶ × 1.6 × 10^-19 J

∴ r₀ = `(9xx10^9xx(92xx1.6xx10^-19)(2xx1.6xx10^-19))/(5xx10^6xx1.6xx10^-19)`

⇒ r = 5.2 × 10^-14

m = 5.3 × 10^-12 cm.