1. Energy of the first photon

A hydrogen atom in ground state has ionization energy $13.6\,\text{eV}.$

The atom absorbs a photon of frequency $\nu_1$ and the electron is ejected with kinetic energy $10\,\text{eV}$.

So, by energy conservation, $h\nu_1 = 13.6 + 10 = 23.6\,\text{eV}.$

2. Formation of positronium

Now the emitted electron (kinetic energy $10\,\text{eV}$) combines with a positron initially at rest.

A positronium atom is formed in its ground state and a photon of frequency $\nu_2$ is emitted. The center of mass (CM) of the positronium moves with kinetic energy $5\,\text{eV}$.

We need the binding energy of positronium in ground state.

3. Ground-state energy of positronium

For a Bohr atom, $E_n = -\frac{\mu}{m_e}\frac{13.6}{n^2}\,\text{eV},$ where $\mu$ is the reduced mass.

For positronium, electron and positron have equal masses $m_e$, so $\mu = \frac{m_e\,m_e}{m_e+m_e} = \frac{m_e}{2}.$

Hence ground-state energy is $E_1 = -\frac{1}{2}(13.6) = -6.8\,\text{eV}.$

So the binding energy released in forming ground-state positronium is $6.8\,\text{eV}.$

4. Energy available before and after positronium formation

Before formation:

• electron KE = $10\,\text{eV}$
• positron KE = $0$
• no binding yet

After formation:

• positronium CM kinetic energy = $5\,\text{eV}$
• emitted photon energy = $h\nu_2$
• internal energy decreases by $6.8\,\text{eV}$ due to binding

Thus total released/available energy is $10 + 6.8 = 16.8\,\text{eV}.$

This is shared between emitted photon and CM kinetic energy of positronium: $h\nu_2 + 5 = 16.8.$ Therefore, $h\nu_2 = 11.8\,\text{eV}.$

5. Difference between the two photon energies

We need $h\nu_1 - h\nu_2 = 23.6 - 11.8 = 11.8\,\text{eV}.$

So the required difference is $\boxed{11.8\,\text{eV}}.$

6. Comparison with stored answer

Stored correct answer: $11.7$ to $11.9$

Our derived answer is $11.8$, which lies in the given range. Hence it agrees.