Question

The binding energy of the ground state of positronium is a factor f times that of a hydrogen atom. f =

• A. 1
• B. $\frac{1}{2}$
• C. 2
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{1}{2}$

Answer 2

An electron moving in the field of a proton or positron has potential energy –$\frac{e^{2}}{\tau}$, for which the radial Schrodinger equation gives eigenvalue

E_n = –$\frac{\mu e^{2}}{2\hslash^{2}n^{2}}$, n = 1, 2, 3, …..,

where µ is the reduced mass of the orbiting electron. For the positronium µ =$\frac{1}{2}$m_e, for hydron atom µ » m_e. Hence

f =$\frac{E_{p}}{E_{h}}$=$\frac{\frac{1}{2}m_{e}}{m_{e}}$=$\frac{1}{2}$.