JEE Advanced 2025 Paper 2 Physics Question 10 Solution

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency $\nu_1$ 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 $\nu_2$ . 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 ____

Step 1: Calculate the energy of the first photon ( $h\nu_1$ ). In the hydrogen atom, the ionization energy from ground state is 13.6 eV. $h\nu_1 = E_{ionisation} + K_e = 13.6 \text{ eV} + 10 \text{ eV} = 23.6 \text{ eV}$

Step 2: Determine the energy levels of a positronium atom. Positronium consists of an electron and a positron. The reduced mass is $\mu = \frac{m_e m_e}{m_e + m_e} = \frac{1}{2}m_e$ . Since energy $E \propto \mu$ , the ground state energy of positronium is half that of Hydrogen: $E_{gs,pos} = \frac{1}{2} (-13.6 \text{ eV}) = -6.8 \text{ eV}$

Step 3: Use conservation of energy for the second process. Initial energy (electron kinetic energy + positron energy) equals final energy (positronium energy + photon $h\nu_2$ ): $K_e + K_p = (E_{gs,pos} + K_{com}) + h\nu_2$ $10 \text{ eV} + 0 \text{ eV} = -6.8 \text{ eV} + 5 \text{ eV} + h\nu_2$ $10 = -1.8 + h\nu_2 \implies h\nu_2 = 11.8 \text{ eV}$

Step 4: Calculate the difference between the two photon energies: $\Delta E = |h\nu_1 - h\nu_2| = |23.6 - 11.8| = 11.8 \text{ eV}$ Final Answer is 11.8.

Question 10

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