Question

Positronium is like a H atom with the proton replaced by positron (a positively charged antiparticle of the electron which is as massive as electron). The ground state energy of positronium would be

• A. $- 3.4\text{ eV}$
• B. $- 5.2\text{ eV}$
• C. $- 6.8\text{ eV}$
• D. $- 10.2\text{ eV}$

Answer 1

Answer: (C)

$- 6.8\text{ eV}$

Answer 2

Bohr’s formula for ground state energy

$E = - \frac{me^{4}}{8\varepsilon_{0}^{2}h^{2}}$ $(\because n = 1).........(1)$

Here, m is reduced mass of electron and positron in positronium

$\therefore m = \frac{m_{e}m_{p}}{m_{e} + m_{p}} = \frac{m_{e}}{2}$ $(\because m_{e} = m_{p})$

$\therefore$ Ground state energy of positronium

$E = - \frac{\left( \frac{m_{e}}{2} \right)e^{4}}{8\varepsilon_{0}^{2}h^{2}} = - \frac{1}{2}\left( \frac{m_{e}e^{4}}{8\varepsilon_{0}^{2}h^{2}} \right)$ (using(i)

$= - \frac{1}{2} \times 13.6eV$ $\left\lbrack \frac{m_{e}e^{4}}{8\varepsilon_{0}^{2}h^{2}} = 13.6eV \right\rbrack$

$= - 6.8eV.$