JEE Advanced 2025 Paper 2 · Q10 · Bohr Model

Question

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency $
u_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 $
u_2$. The center of mass of the resulting positronium atom moves with a kinetic energy of $5$ eV. 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, the difference between the two photon energies (in eV) is __________.

SolveKar AI · Accepted Answer

Step 1 (ionization + KE): $h
u_1 = E_{	ext{ion}}(H) + KE_e = 13.6 + 10 = 23.6$ eV. Step 2 (positronium formation): the moving electron ($10$ eV KE) collides with a positron at rest. Reduced-mass scaling for a Bohr atom with two equal masses gives ground-state binding energy $= 13.6/2 = 6.8$ eV. Lab-frame energy conservation: $KE_{	ext{initial}} = KE_{	ext{COM}}(	ext{positronium}) + 	ext{emitted photon energy} + (-	ext{binding})$. Conservation: $10$ eV (initial KE of electron) $= 5$ eV (KE of COM positronium) $+ h
u_2 - 6.8$ eV. Rearranging: $h
u_2 = 10 - 5 + 6.8 = 11.8$ eV. Difference $h
u_1 - h
u_2 = 23.6 - 11.8 = 11.8$ eV.

JEE Advanced 2025 Paper 2 · Q10 · Bohr Model

Solution