JEE Main 2024 Physics 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 is × 10⁵ m/s approximately. ( = 9 × 10⁹ SI unit, mass of α particle = 6.72 × 10⁻²⁷ kg)
Answer: 156
Solution
½mv² = k(2e)(80e)/r ⇒ v = √(2 × 9 × 10⁹ × 160 × (1.6 × 10⁻¹⁹)²/(4.5 × 10⁻¹⁴ × 6.72 × 10⁻²⁷)) = 1.56 × 10⁷ m/s = 156 × 10⁵ m/s.