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← JEE Advanced 2025 · Paper 2
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JEE Advanced 2025 · Paper 2, Physics question 10

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency ν1\nu_1 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 ν2\nu_2. 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 and worked solution

Official answer: 11.7

First photon. It must ionise hydrogen and supply the electron's kinetic energy:
E1=13.6+10=23.6 eV.E_1=13.6+10=23.6\ \text{eV}.

Positronium. With reduced mass m2\frac m2, the ground-state energy is −13.62=−6.8-\frac{13.6}{2}=-6.8 eV.

Second photon. Energy is conserved, with the photon's momentum negligible. The positronium, of mass 2m2m, has kinetic energy 5 eV, consistent with momentum conservation:
10=5−6.8+E2  ⟹  E2=11.8 eV.10=5-6.8+E_2\implies E_2=11.8\ \text{eV}.

Difference. E1−E2=11.8E_1-E_2=11.8 eV.