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 ______.

The ground-state energy of a hydrogen atom is $$E_{H} = -13.6 \,{\rm eV}$$. To ionise hydrogen and simultaneously give the liberated electron a kinetic energy of $$10 \,{\rm eV}$$, the absorbed photon must supply

$$h\nu_1 = 13.6 \,{\rm eV} + 10 \,{\rm eV} = 23.6 \,{\rm eV}$$

(The recoil of the heavy proton is neglected because its share of energy is < 0.01 eV.)

Next, the free electron (kinetic energy $$10 \,{\rm eV}$$) meets a positron that is initially at rest. They form a positronium atom and emit a photon of frequency $$\nu_2$$. Positronium can be treated as a Bohr atom whose reduced mass is $$\mu = m_e/2$$, so every energy level is exactly half that of the corresponding hydrogen level. Hence the ground-state energy of positronium is

$$E_{\text{Ps}} = \tfrac{1}{2}( -13.6 \,{\rm eV}) = -6.8 \,{\rm eV}$$

During this formation the centre of mass of the positronium acquires a kinetic energy of $$5 \,{\rm eV}$$. Apply energy conservation between the instant just before recombination and the instant just after the photon emission:

Initial energy (electron only): $$10 \,{\rm eV}$$

Final energy: $$h\nu_2 + 5 \,{\rm eV} + (-6.8 \,{\rm eV})$$

Setting them equal,

$$10 = h\nu_2 + 5 - 6.8$$ $$\Rightarrow\; h\nu_2 = 10 - 5 + 6.8 = 11.8 \,{\rm eV}$$

The difference between the two photon energies is therefore

$$h\nu_1 - h\nu_2 = 23.6 \,{\rm eV} - 11.8 \,{\rm eV} = 11.8 \,{\rm eV}$$

Hence the required energy difference lies in the range $$11.7\;{\rm eV}\;-\;11.9\;{\rm eV}$$.