A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency v₁ 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 v₂. 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 correct answer is 11.8.

Let us break down the problem into step-by-step physical processes and apply the conservation of energy to find the difference between the two photon energies.

Step 1: Energy of the first photon ($E_1$)
The hydrogen atom is initially at rest in its ground state. The ground state energy of a hydrogen atom is:
$E_{1,\text{H}}=-13.6\text{eV}$
To eject the electron, the energy equal to the ionization energy of hydrogen (13.6 eV) must be supplied. The electron is ejected with a kinetic energy of $K_e=10\text{eV}$.
By conservation of energy, the energy of the absorbed photon of frequency $v_1$ is:
$E_1=h{v}_1=13.6\text{eV}+10\text{eV}=23.6\text{eV}$

Step 2: Ground state energy of the positronium atom
A positronium atom consists of an electron (mass $m$) and a positron (mass $m$) orbiting around their common center of mass. The reduced mass ($\mu$) of this system is:
$\mu =\frac{m·m}{m+m}=\frac{m}{2}$
Since the energy levels in a Bohr atom are directly proportional to the reduced mass, the ground state energy of the positronium atom ($E_{1,\text{ps}}$) is related to that of the hydrogen atom by:
$E_{1,\text{ps}}=\frac{\mu }{m}{E}_{1,\text{H}}=\frac{1}{2}(-13.6\text{eV})=-6.8\text{eV}$

Step 3: Energy of the second photon ($E_2$)
Next, the ejected electron (with kinetic energy $K_e=10\text{eV}$) combines with a positron at rest (kinetic energy $0\text{eV}$) to form a positronium atom in its ground state. The center of mass of the resulting positronium atom moves with a kinetic energy of $K_{\text{ps}}=5\text{eV}$, and a photon of frequency $v_2$ is emitted.
Applying the law of conservation of energy (excluding the rest mass energy, which is conserved separately):
$K_e=E_{1,\text{ps}}+K_{\text{ps}}+E_2$
Substitute the values into the equation:
$10\text{eV}=-6.8\text{eV}+5\text{eV}+E_2$
$10\text{eV}=-1.8\text{eV}+E_2$
$E_2=10\text{eV}+1.8\text{eV}=11.8\text{eV}$

Step 4: Difference between the two photon energies
The difference between the two photon energies is:
$|E_1-E_2|=23.6\text{eV}-11.8\text{eV}=11.8\text{eV}$