An α -particle of energy 5 MeV 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 MeV is scattered through 180° by a stationary uranium nucleus. The distance of closest approach is of the order (A) 1Å (B) 10- 10 cm (C) 10- 12 cm (D) 10- 15 cm

Tardigrade Admin · Accepted Answer

According to work energy theorem KE=PE KE=(1/4 π ϵ 0)(q1 q2/r) r=(1/4 π ϵ 0)(92 e ⋅ 2 e/E) r=(9 × 1 09 × 92 × 2 (1.6 × 1 0- 19)2/5 × 1 06 × (1.6 × 1 0- 19)) r=5.3× 10- 14m ≈10- 12 cm.

Q. An $α$ -particle of energy $5 M e V$ is scattered through $18 0^{\circ}$ by a stationary uranium nucleus. The distance of closest approach is of the order

$1Å$

$1 0^{- 10} c m$

$1 0^{- 12} c m$

$1 0^{- 15} c m$

Q. An α -particle of energy 5MeV is scattered through 180∘ by a stationary uranium nucleus. The distance of closest approach is of the order

1Å

10−10cm

10−12cm

10−15cm

According to work energy theorem KE=PE KE=4πϵ0​1​rq1​q2​​ r=4πϵ0​1​E92e⋅2e​ r=5×106×(1.6×10−19)9×109×92×2(1.6×10−19)2​ r=5.3×10−14m≈10−12cm.

Q. An $α$ -particle of energy $5 M e V$ is scattered through $18 0^{\circ}$ by a stationary uranium nucleus. The distance of closest approach is of the order

$1Å$

$1 0^{- 10} c m$

$1 0^{- 12} c m$

$1 0^{- 15} c m$