In a Geiger Marsden experiment, an alpha particle of energy 6 meV hits a nucleus of z = 72. The distance of closest approach nearly is

34.6 fm

The correct answer is Option 1: 34.6 fm

When an alpha particle approaches a nucleus head-on, its initial kinetic energy is completely converted into electrostatic potential energy at the point of closest approach.

Therefore,

K.E = (1 / 4πϵ₀) × (q₁q₂ / r₀)

For the given case:

Charge of alpha particle = 2e
Charge of nucleus = Ze

Thus,

K.E = (1 / 4πϵ₀) × (2Ze² / r₀)

Rearranging,

r₀ = (1 / 4πϵ₀) × (2Ze² / K.E)

Substituting the values:

(1 / 4πϵ₀) = 9 × 10⁹
Z = 72
e = 1.6 × 10⁻¹⁹ C
K.E = 6 MeV = 6 × 10⁶ × 1.6 × 10⁻¹⁹ J

r₀ = (9 × 10⁹ × 2 × 72 × (1.6 × 10⁻¹⁹)²) / (6 × 10⁶ × 1.6 × 10⁻¹⁹)

Simplifying,

r₀ = (9 × 10⁹ × 144 × 1.6 × 10⁻¹⁹) / (6 × 10⁶)

r₀ = 3.456 × 10⁻¹⁴ m

r₀ ≈ 34.6 fm.

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