Question

If muonic hydrogen atom is an atom in which a negatively charged muon ($\mu$) of mass about $207 m_e$ revolves around a proton, then first Bohr radius of this atom is $(r_e = 0.53 \times 10^{-10} m)$, if ground state energy of electron is $-13.6\, eV$ then what is the ground state energy of muonic hydrogen atom?

• A. $1.8\,keV$
• B. $-2.8\,keV$
• C. $-3.8\,keV$
• D. $4.8\,keV$

Answer 1

Answer: (B)

According to Bohr’s atomic model energy,
$E \propto m $
$ \therefore \frac{E_{\mu}}{E_{e}} = \frac{m_{\mu}}{m_{e}} $
Here, $E_{e} = -13.6 eV ; $
$ m_{\mu} = 207 m_{e} $
$\therefore E_{\mu} =\frac{m_{\mu}}{m_{e}}E_{e} $
$ = \frac{207m_{e}}{m_{e} } \times\left(-13.6\right)eV $
$ = -2815.2\, eV$
$= -2.8\, keV $