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

Answer 1

Answer: (C)

Bohr’s formula for ground state energy,
$E=-\frac{m{e}^4}{8{\epsilon }_0^2{h}^2}\left(\because n=1\right)...\left(i\right)$
Here, $m$ is reduced mass of electron and positron in positronium
$\therefore m=\frac{m_e{m}_p}{m_e+m_p}=\frac{m_e}{2}\left(\because m_e=m_p\right)$
$\therefore$ Ground state energy of positronium
$E=-\frac{\left(\frac{m_e}{2}\right)e^4}{8{\epsilon }_0^2{h}^2}$
$=-\frac{1}{2}\left(\frac{m_e{e}^4}{8{\epsilon }_0{h}^2}\right)$
$=-\frac{1}{2}\times 13.6eV\left[\frac{m_e{e}^4}{8{\epsilon }_0^2{h}^2}=13.6eV\right]$
$=-6.8eV$