If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach i $s ________$ m.
(Atomic number of gold = 79 and 1 ────── 4πε₀ = 9 × 10⁹ in SI units)

Question

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach i $s ________$ m.

(Atomic number of gold = 79 and 1 ────── 4πε₀ = 9 × 10⁹ in SI units)

G1 Physics · Accepted Answer

For a head-on approach, the alpha particle momentarily comes to rest at the distance of closest approach. Therefore its initial kinetic energy is converted into electrostatic potential energy. Initial energy: K = 7.7 MeV = 7.7 × 10⁶ × 1.6 × 10⁻¹⁹ J. The alpha-particle charge is qα = +2e. The gold-nucleus charge is q₍Au₎ = +79e. Hence, K = 1 ────── 4πε₀ × (2e)(79e) ────────── r . Substitute 1 ────── 4πε₀ = 9 × 10⁹ and e = 1.6 × 10⁻¹⁹ C. Solving for r gives r ≈ 2.95 × 10⁻¹⁴ m. Correct option: A.

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach i s ________ m. (Atomic number of gold = 79 and 1 ────── 4πε₀ = 9 × 10⁹ in SI units)