JEE Advanced 2025 — Paper 2 — PHY-III solution

JEE Advanced2025Paper 2PHY-III

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency $ν_{1}$ 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 $ν_{2}$ . 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 ____

Your answer

Solution

Step 1 (absorption by H): Ground-state H needs 13.6 eV to ionize, and the ejected electron carries 10 eV kinetic energy. Energy conservation:

$h ν_{1} = 13.6 + 10 = 23.6$ eV

Step 2 (positronium formation): Positronium is a Bohr-like atom with reduced mass $μ = m_{e} /2$ . Its ground-state binding energy is

$E_{1}^{Ps} = - 13.6 \cdot \frac{μ}{m _{e}} = - 6.8$ eV

So 6.8 eV must be released when the electron (10 eV kinetic energy) binds with the positron (at rest). The positronium itself carries 5 eV as kinetic energy of its centre of mass. The remaining energy escapes as the photon $h ν_{2}$ :

$h ν_{2} = (10 - 5) + 6.8 = 11.8$ eV

Difference: $h ν_{1} - h ν_{2} = 23.6 - 11.8 = 11.80$ eV.

← Back to full paper

Practice more PHY-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →