Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium ?

Explanation

According to Bohr atomic model , energy of `n^(th)` orbit is given by
`E_n=(-me^4)/(8epsilon_0^2 n^2h^2)`
where m is the reduced mass.
In case of original atom, we know that nucleus is heavy in comparison to electron and hence nucleus can be assumed to be at rest and just electron is revolving around it. But when positron is used in place of proton then both the particles (electron and positron) are having the same mass and they will be orbiting around their centre of mass which is at the centre of line joining them. In such a two-particle system, if we assume positron to be at rest then mass of the electron to be used will be reduced mass. Reduced mass of the system is `m=m_1m_2//(m_1+m_2)`, hence in this case reduced mass becomes `m_e//2`.
For H-atom , `m=m_e`
For positronium , `m=m_e//2`
The ground state energy (n=1) for positronium is thus given by
`E_n=(-(m_e//2)e^4)/(8epsilon_0^2xx(1^2)xxh^2)=-13.6/2`eV=-6.8eV (`because (m_ee^4)/(8epsilon_0^2 n^2 h^2)`=13.6eV)