A particle of mass 200 MeV/c² collides with a hydrogen atom at rest. Soon after the collision the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is
N/4. The value of N is :
(Given the mass of the hydrogen atom to be 1 GeV/c²) ______ .

Solution

Physics sir · 4 steps

Momentum passes fully to the atom, so its recoil takes a fixed share of the KE; the rest excites it.

1. Momentum is conserved. The particle stops, so all its momentum p goes to the hydrogen atom. Write each kinetic energy as K = p²/2m with the same p.

   Kₐₜₒₘ = p²/2M = K m/M = K × (200 MeV/c²)/(1000 MeV/c²) = K/5

2. Energy lost. The kinetic energy that disappears is K - K/5 = 4K/5. It goes into lifting the electron from n = 1 to n = 2, which needs 13.6(1 - 1/4) = 10.2 eV.
3. Solve for K. 4K/5 = 10.2 eV gives K = 12.75 eV = 51/4 eV.
4. So. Comparing with N/4, N = 51.

The common slip: 40.8. Setting the whole initial kinetic energy equal to 10.2 eV ignores the atom's recoil and gives N = 40.8, not 51.

The idea to keep. In an inelastic collision, conserve momentum first. Then the kinetic energy lost is what goes into excitation, and the recoiling body always keeps part of the KE.

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