Positronium is a bound state of an electron and a positron (a...

In a two-body system where the nucleus is not infinitely massive, the effective mass of the orbiting particle is replaced by the reduced mass $\mu$ of the system.

For positronium, both particles have mass $m_e$. The reduced mass is:
$\mu = \dfrac{m_e \cdot m_e}{m_e + m_e} = \dfrac{m_e}{2}$

The energy of a Bohr orbit is directly proportional to the reduced mass ($E \propto \mu$). Therefore, the ground state energy of positronium is:
$E = E_H \times \left(\dfrac{\mu}{m_e}\right) = -13.6 \text{ eV} \times \dfrac{1}{2} = -6.8 \text{ eV}$

The radius of a Bohr orbit is inversely proportional to the reduced mass ($r \propto \dfrac{1}{\mu}$). Therefore, the first Bohr radius of positronium is:
$r = r_H \times \left(\dfrac{m_e}{\mu}\right) = a_0 \times 2 = 2a_0$.

Answer: $-6.8 \text{ eV}, 2a_0$