Question

Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon $\left({\mu }^{-}\right)$ of mass about $207m_e$ orbits around a proton].

Answer 1

Step 1: Find first Bohr’s radius of muonic hydrogen atom.
Formula used:$r=\frac{n^2{h}^2}{4{\pi }^{2}mk{e}^2}$
Given,
mass of negatively charged muon,
$m_{\mu }=207m_e$
Radius of first Bohr orbit,
$r_e=0.53\times {10}^{-10}m$
As we know, Bohr radius, $r=\frac{n^2{h}^2}{4{\pi }^{2}mk{e}^2}$
For ground state, $r\propto \frac{1}{m}$
radius of ground state H-atom,
$r_e\propto \frac{1}{m_e}$
Let $r_{\mu }$ be the radius of muonic hydrogen atom.
radius of ground state muonic H-atom,
$r_{\mu }\propto \frac{1}{m_{\mu }}$
At equilibrium, $m_{\mu }{r}_{\mu }=m_e{r}_e$
$207m_e{r}_{\mu }=m_e\times (0.53\times {10}^{-10})$
$r_{\mu }=2.56\times {10}^{-13}m$
Step 2: Find ground state energy of muonic hydrogen atom.
Formula used: $E_n=-\frac{m{e}^4}{8n^2{ϵ}_0^2{h}^2}$
Given,
mass of negatively charged muoun,
$m_{\mu }=207m_e$
Ground state energy for H - atom,
$E_e=-13.6eV$
As we know, energy in $n^{th}$ state,
$E_n=-\frac{m{e}^4}{8n^2{ϵ}_0^2{h}^2}$
Energy of ground state H-atom,
$E_e\propto m_e$
Energy of ground state muonic H-atom,
$E_{\mu }\propto m_{\mu }$
Taking the ratio of the energies
$\frac{E_e}{E_{\mu }}=\frac{m_e}{m_{\mu }}$
$\frac{E_e}{E_{\mu }}=\frac{m_e}{207m_e}$
$E_{\mu }=207E_e$
$E_{\mu }=207(-13.6)=-2.81keV$
Final answer: $2.56\times {10}^{-13}m;-2.81keV$