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. $-3.4\,eV$
• B. $-5.2\,eV$
• C. $-6.8\,eV$
• D. $-10.2\,eV$

Answer 1

Answer: (C)

Bohr’s formula for ground state energy,
$E = -\frac{me^{4}}{8\varepsilon_{0}^{2}h^{2}} \quad\left(\because n=1\right)...\left(i\right) $
Here, $m$ is reduced mass of electron and positron in positronium
$\therefore m = \frac{m_{e}m_{p}}{m_{e}+m_{p}} = \frac{m_{e}}{2} \quad\left(\because m_{e} = m_{p}\right)$
$\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}h^{2}}\right) $
$ = -\frac{1}{2}\times 13.6 \,eV \quad\left[\frac{m_{e} e^{4}}{8\varepsilon_{0}^{2}h^{2}} = 13.6 eV\right] $
$ = -6.8\, eV$