In This post you will get answer of Q.
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
Solution:
Bohr’s formula for ground state energy,
$E = -frac{me^{4}}{8varepsilon_{0}^{2}h^{2}} quadleft(because n=1right)…left(iright) $
Here, $m$ is reduced mass of electron and positron in positronium
$therefore m = frac{m_{e}m_{p}}{m_{e}+m_{p}} = frac{m_{e}}{2} quadleft(because m_{e} = m_{p}right)$
$therefore $ Ground state energy of positronium
$E= – frac{left(frac{m_{e}}{2}right)e^{4}}{8varepsilon_{0}^{2}h^{2}} $
$= – frac{1}{2} left(frac{m_{e}e^{4}}{8varepsilon_{0}h^{2}}right) $
$ = -frac{1}{2}times 13.6 ,eV quadleft[frac{m_{e} e^{4}}{8varepsilon_{0}^{2}h^{2}} = 13.6 eVright] $
$ = -6.8, eV$
Solution:
Bohr’s formula for ground state energy,
$E = -frac{me^{4}}{8varepsilon_{0}^{2}h^{2}} quadleft(because n=1right)…left(iright) $
Here, $m$ is reduced mass of electron and positron in positronium
$therefore m = frac{m_{e}m_{p}}{m_{e}+m_{p}} = frac{m_{e}}{2} quadleft(because m_{e} = m_{p}right)$
$therefore $ Ground state energy of positronium
$E= – frac{left(frac{m_{e}}{2}right)e^{4}}{8varepsilon_{0}^{2}h^{2}} $
$= – frac{1}{2} left(frac{m_{e}e^{4}}{8varepsilon_{0}h^{2}}right) $
$ = -frac{1}{2}times 13.6 ,eV quadleft[frac{m_{e} e^{4}}{8varepsilon_{0}^{2}h^{2}} = 13.6 eVright] $
$ = -6.8, eV$