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JEE Advance - Physics (2025 - Paper 2 Online - No. 10)

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency $v_1$ 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 $v_2$. 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
11.80

Explanation

Ionization of H (ground state at –13.6 eV)

• Photon of energy $E_1=h\nu_1$ ionizes H and leaves the electron with 10 eV.

• Energy conservation (neglecting proton recoil):

$ E_1 = 13.6\;\text{eV} \;+\;10\;\text{eV} =23.6\;\text{eV}. $

Formation of positronium (Ps)

• Positron initially at rest, electron arrives with 10 eV.

• Ps ground‐state binding energy is 6.8 eV (half of hydrogen’s 13.6 eV).

• The newly formed Ps moves with 5 eV of center‐of‐mass (COM) kinetic energy and emits a photon of energy $E_2=h\nu_2$.

• Writing energy conservation (zero energy set at free e⁺ + e⁻ at rest):

$ \underbrace{10\;\text{eV}}_{\text{initial KE}} =\;\underbrace{(-6.8\;\text{eV})}_{\text{binding}} \;+\;\underbrace{5\;\text{eV}}_{\text{COM KE}} \;+\;E_2 $

so

$ E_2 = 10 -5 +6.8 = 11.8\;\text{eV}. $

Difference between the two photon energies

$ E_1 - E_2 = 23.6\;\text{eV} \;-\;11.8\;\text{eV} = 11.8\;\text{eV}. $

Answer: The difference $h\nu_1 - h\nu_2$ is 11.8 eV.

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