question_answer

Obtain the first Bohr's radius and the ground state energy of a muonic H-atom (i.e. an atom in which a negatively charged muon \[(\overline{\mu })\] of mass about 207 \[{{m}_{e}}\](orbits around a proton).

Answer:

            Muonic hydrogen is the atom in which a negatively charged muon of mass about 207 \[{{m}_{e}}\] revolves round a proton. In Bohr's atom model,     \[r\propto \frac{1}{m}\] \[\because \]       \[\frac{{{r}_{muon}}}{{{r}_{electron}}}=\frac{{{m}_{e}}}{{{m}_{\mu }}}=\frac{{{m}_{e}}}{207{{m}_{e}}}=\frac{1}{207}\]                                     \[[\because {{m}_{\mu }}=207\,{{m}_{e}}]\] Here, \[{{r}_{e}}\] is radius of orbit of electron in H-atom\[=\text{ }0.53\text{ }\overset{{}^\circ }{\mathop{A}}\,\] \[{{r}_{\mu }}=\frac{{{r}_{e}}}{207}=\frac{0.53\times {{10}^{-10}}}{207}=2.56\times {{10}^{-13}}m\] Again in Bohr's atomic model, \[\because \]       \[E\propto m\] \[\therefore \]      \[\frac{{{E}_{\mu }}}{{{E}_{e}}}=\frac{{{m}_{\mu }}}{{{m}_{e}}}=\frac{207\,{{m}_{e}}}{{{m}_{e}}}\] \[\Rightarrow \]   \[{{E}_{\mu }}=207\,{{E}_{e}}\] For ground state, energy of electron in H-atom,             \[{{E}_{e}}=-13.6\,eV\] \[\therefore \]\[{{E}_{\mu }}=207(-13.6)=-2815.2\,eV=-2.8152\,keV\]