In an alpha particle scattering experiment distance of closest approach for the α particle is 4.5 × 10⁻¹⁴ m. If target nucleus has atomic number 80, then maximum velocity of α-particle i s _______ × 10⁵ m/s approximately. (1/(4πε₀) = 9 × 10⁹ SI unit, mass of α particle = 6.72 × 10⁻²⁷ kg)
Physics · Atomic Structure and X-rays · atomic structre and bohr atomic model
JEE Mains 2024 08 Apr Morning Shift

Question

In an alpha particle scattering experiment distance of closest approach for the α particle is 4.5 × 10⁻¹⁴ m. If target nucleus has atomic number 80, then maximum velocity of α-particle i s _______ × 10⁵ m/s approximately. (1/(4πε₀) = 9 × 10⁹ SI unit, mass of α particle = 6.72 × 10⁻²⁷ kg)

Physics · Atomic Structure and X-rays · atomic structre and bohr atomic model

JEE Mains 2024 08 Apr Morning Shift

G1 Physics · Accepted Answer

At the distance of closest approach, initial kinetic energy converts into electrostatic potential energy. For an α-particle of charge +2e approaching a nucleus of charge +Ze: ½mv² = k(2e)(Ze)/r(min) Therefore: v² = 4kZe²/[m r(min)] Using: k = 9 × 10⁹ Z = 80 e = 1.6 × 10⁻¹⁹ C m = 6.72 × 10⁻²⁷ kg r(min) = 4.5 × 10⁻¹⁴ m we get: v ≈ 1.56 × 10⁷ m/s = 156 × 10⁵ m/s Hence: answer = 156