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.
(1/(4 π ε₀)=9 × 10⁹ SI unit, mass of α particle =6.72 × 10⁻²⁷ kg)

Solution

Physics sir · 4 steps

At closest approach the alpha particle stops, so all its kinetic energy has become Coulomb energy.

1. Energy conservation. In a head-on approach the alpha particle slows until it stops at the distance of closest approach r₀. There its whole initial kinetic energy has turned into electrostatic potential energy between its charge 2e and the nuclear charge Ze.

   1/2mv² = 1/4πε₀ (2e)(Ze)/r₀

2. Potential energy. (9 × 10⁹ × 2 × 80 × (1.6 × 10⁻¹⁹)²)/(4.5 × 10⁻¹⁴) = (3.6864 × 10⁻²⁶)/(4.5 × 10⁻¹⁴) = 8.192 × 10⁻¹³ J.
3. Solve for v. v² = (2 × 8.192 × 10⁻¹³)/(6.72 × 10⁻²⁷) = 2.438 × 10¹⁴ m² s⁻². So v = 1.56 × 10⁷ m s⁻¹.
4. So. v = 156 × 10⁵ m s⁻¹, so the answer is 156.

The common slip: 110. Taking the alpha particle's charge as e instead of 2e halves the energy and gives about 110, not 156.

The idea to keep. Distance of closest approach means the particle stops for an instant: set 1/2mv² equal to k(2e)(Ze)/r₀. Always give the alpha particle a charge of 2e.

Answer from the NTA final answer key. Solution by PrepWiser.