Question

An α − α−particle of energy 5 mev 5mev is scattered through 180 ∘ 180∘ by a fixed uranium nucleus. The distance of the closest approach is of the order of

Answer 1

Answer:

Distance of closet approx:−

r0=4π∈0Eze2e

KE=PE

21mv2=4π∈01rq1q2

r=5×106×1.6×10−199×109×2×92×(1.6×10−19)2

r=5.3×10−14m≈10−12cm