7.9 MeV α-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately)________m.
[1/(4πε₀) = 9 × 10⁹ Nm²/C² and electron charge = 1.6 × 10⁻¹⁹ C]

Question

7.9 MeV α-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately)________m.

[1/(4πε₀) = 9 × 10⁹ Nm²/C² and electron charge = 1.6 × 10⁻¹⁹ C]

G1 Physics · Accepted Answer

For a head-on Rutherford encounter, the α-particle momentarily comes to rest at the distance of closest approach. Hence its initial kinetic energy is converted into electrostatic potential energy. The α-particle charge is q₁ = +2e. The gold-like target nucleus has charge q₂ = +Ze = +79e. Therefore, K = kq₁q₂/r, where k = 1/(4πε₀). Substitute the data: 7.9 × 10⁶ × 1.6 × 10⁻¹⁹ = [9 × 10⁹ × 2 × 79 × (1.6 × 10⁻¹⁹)²]/r. Hence, r = 2.88 × 10⁻¹⁴ m. This is the estimated radius corresponding to the dis…