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

Question

An α -particle of energy 5 M e V is scattered through 180 ° by a stationary uranium nucleus. The distance of closest approach is of the order, 1 Å, 1 0 - 10 c m, 1 0 - 12 c m, 1 0 - 15 c m

MARKS App · Accepted Answer

According to work energy theorem K E = P E K E = 1 4 π ε 0 q 1 q 2 r r = 1 4 π ε 0 92 e ⋅ 2 e E r = 9 × 1 0 9 × 92 × 2 1.6 × 1 0 - 19 2 5 × 1 0 6 × 1.6 × 1 0 - 19 r = 5.3 × 1 0 - 14 m ⁡ ≈ 1 0 - 12 c m .

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