A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency and ejects the electron with a kinetic energy of 10 eV. The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency . The center of mass of the resulting positronium atom moves with a kinetic energy of 5 eV. It is given that positron has the same mass as that of electron and the positronium atom can be considered as a Bohr atom, in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process, the difference between the two photon energies (in eV) is ____
Answer: 11.8
Solution
Photon 1: Ionizes hydrogen and gives electron 10 eV kinetic energy:
Positronium formation: Ground state energy of positronium (reduced mass = m/2):
Energy released in formation = 6.8 eV. The electron had 10 eV kinetic energy; after forming positronium with 5 eV COM kinetic energy, the energy available for photon emission is:
Difference:
Positronium formation: Ground state energy of positronium (reduced mass = m/2):
Energy released in formation = 6.8 eV. The electron had 10 eV kinetic energy; after forming positronium with 5 eV COM kinetic energy, the energy available for photon emission is:
Difference:
JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer