Physics - Modern Physics Question with Solution | TestHub

For positronium reduced mass $=\frac{\mathrm{mm}}{\mathrm{m}+\mathrm{m}}$ $=\frac{{\mathrm{m}}^2}{2\mathrm{m}}=\frac{\mathrm{m}}{2}$ Hence Rydberg's constant for positronium is half of that for hydrogen atom. Hence, ground state energy for positronium $=-\frac{13.6}{2}=-6.8\mathrm{eV}$