Question

Exercise 81 obtained formulas for hydrogen like atoms in which the nucleus is not assumed infinite, as in the chapter, but is of mass$m_1$, while$m_2$is the mass of the orbiting negative charge. In positronium, an electron orbits a single positive charge, as in hydrogen, but one whose mass is the same as that of the electron -- a positron. Obtain numerical values of the ground state energy and “Bohr radius” of positronium.

Short answer

(a) The ground State energy of positronium is -6.8 eV .

(b) Bohr radius of positronium is 0.106 nm .

Step-by-step solution

Energy of ground-state of positronium

As you know from that, the ground state energy is,
$E_{ground}=\frac{Z^2\mu }{m}\frac{E_1}{n^2}$

Where, z is the atomic number, $\mu$is the Reduced mass, n is the principal quantum number, m is the mass, $E_1$is the Energy of ground state of hydrogen atom.

You also know that the electron and positron have same masses.

Hence, the reduced mass will be half of the mass of electron
$$\begin{gathered} E_{ground}=\frac{Z^2\mu }{m}\frac{E_1}{n^2} \\ =\frac{1^2{\displaystyle \frac{1}{2}}m}{m}\frac{E_1}{1^2} \\ =-6.8eV \end{gathered}$$

Hence, ground state energy of the positronium is -6.8 eV .

Bohr Radius of the positronium

As you know that,

The Bohr’s Radius
$r_n=\frac{m}{z\mu }{a}_0$

Where, $a_0$is radius of hydrogen atom

If the reduced mass will be half of the mass of electron,

$$\begin{gathered} r_n=\frac{m}{Z\mu }{a}_0 \\ =\frac{m}{Z{\displaystyle \frac{1}{2}}m}{a}_0 \\ =0.106nm \end{gathered}$$

Hence, Bohr Radius of the positronium is 0.106 nm .