Answer:
v = 0.525 /√r
Explanation:
For this problem we must use Newton's second law where force is electric given by Coulomb's law.
F = m a
k q₁ q₂ / x² = m dv / dt
Where index 1 we will use it for the proton and index 2 for the alpha particle
k / m q₁q₂ / x₂ = dv/dx dx/dt = dv/dx v
(k / m q₁q₂) dx / x² = v dv
We integrate
(k / m q₁q₂) (-1 / x) = v² / 2
We evaluate between the lower limit x = r and v = 0 to the upper limit x = 2r
(k / mq₁q₂) (-1 / 2r + 1 / r) = ½ (v²-0)
v² = 2k/m q₁q₂ (1 / 2r)
q₁ = e
q₂ = 2e
v² = k 2e² /m 1 /r
v² = 2k e²/m 1 /r
Let's replace the values
v² = 2 8.99 10⁹ (1.6 10⁻¹⁹)² /1.67 10⁻²⁷ 1 / r
v = √( 27.562 10⁻² 1 / r)
v = 5.25 10⁻¹ 1 / √r
v = 0.525 /√r
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