A ground-state hydrogen atom at rest absorbs a photon of frequency ν₁ and ejects an electron with kinetic energy 10 eV. The electron combines with a positron at rest to form ground-state positronium, emitting a photon of frequency ν₂. The positronium center of mass has kinetic energy 5 eV. Find h(ν₁ − ν₂) in eV.
Physics · Circular Motion
JEE Advanced 2025 Paper 2

Question

A ground-state hydrogen atom at rest absorbs a photon of frequency ν₁ and ejects an electron with kinetic energy 10 eV. The electron combines with a positron at rest to form ground-state positronium, emitting a photon of frequency ν₂. The positronium center of mass has kinetic energy 5 eV. Find h(ν₁ − ν₂) in eV.

Physics · Circular Motion

JEE Advanced 2025 Paper 2

G1 Physics · Accepted Answer

Hydrogen ionization from ground state requires 13.6 eV. Therefore hν₁ = 13.6 + 10 = 23.6 eV. Ground-state positronium binding energy is 6.8 eV. For electron + positron: initial kinetic energy = 10 eV. After formation: 10 = 5 − 6.8 + hν₂ ⇒ hν₂ = 11.8 eV. Therefore h(ν₁−ν₂) = 23.6−11.8 = 11.8 eV.