Question

Determine the distance of closest approach when an alpha particle of kinetic energy 4.5 mev strikes a nucleus of z = 80, stops and reverses its direction.

Answer 1

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Answer 2

The distance of closest approach is 5.12×10⁻¹⁴ m.

Let 'r' be the distance of closest approach.

According to Coloumb's Law, we know that the Electrostatic potential energy is given as

U = {[1/(4Пξ₀)] × [(Ze × 2e)/r]}     =  [(2Ze²)/4Пξ₀r]

As Kinetic Energy is being converted to potentail energy, so

U = 4.5 MeV = (4.5×10⁶ × 1.6×10⁻¹⁹)J    [1 MeV = 10⁶ eV and 1 eV = 1.6×10⁻¹⁹J]

1/(4Пξ₀) = 9×10⁹ N m²/C²

Z = 80

So, r = {[1/(4Пξ₀)] × [(Ze × 2e)/U]}

        = [9×10⁹ × 2×80×(1.6×10⁻¹⁹)²]/[4.5×10⁶ × 1.6×10⁻¹⁹]

        = 5.12×10⁻¹⁴ m