The radius of the first permitted Bohr orbit for the electron, in a hydrogen atom equals 0.51 $\stackrel{0}{\mathrm{A}}$ and its ground state energy equals $-$ 13.6 e V. If the electron in the hydrogen atom is replaced by muon $\left({\mu }^{-1}\right)$) [Charge same as electron and mass 207 m_e], the first Bohr radius and ground state energy will be [NEET (Odisha) 2019]

Given, radius of first orbit for electron r₁ = 0.51$\stackrel{0}{\mathrm{A}}$

Ground state energy of electron, E₁ = $-$ 13.6 e V
Mass of electron = m_e
Mass of muon, $m_{\mu }$ = 207m.
Mass of nucleus, M = 1836m_e
When electron in hydrogen atom is replaced by muon, the reduced mass of muon is

$m_{\mu }^{\text{'}}=\frac{m_{\mu }M}{m_{\mu }+M}$           …(i)

Substituting the given values in Eq. (il, we get

$m_{\mu }^{\text{'}}=\frac{207m_e\times 1836m_e}{207m_e+18\beta 6m_e}\approx 186m_e$          …(ii)

The radius of first orbit in hydrogen atom for electron is given by

$r_1=\frac{h^2{\mathit{\epsilon }}_0}{\pi m_e{e}^2}$                      …(iii)

The radius of first orbit for muon is given by

$r_1^{\text{'}}=\frac{h^2{\epsilon }_0}{\pi m_{\mu }^{\text{'}}{e}^2} \left(\because \text{ charge of }\mu =\text{ charge of }{e}^{-}\right)$

$=\frac{h^2{\epsilon }_0}{\pi \times 186m_e{e}^2}$     [from Eq. (ii)]

$=\left(\frac{h^2{\epsilon }_0}{\pi m_e{e}^2}\right)\frac{1}{186}=\frac{r_1}{186}$  [from Eq. (iii)]

$=\frac{0.51}{186} \stackrel{0}{\mathrm{A}}$                    $\left(\because r_1=0.51 \stackrel{0}{\mathrm{A}}\right)$

$=2.74\times {10}^{-13} \mathrm{m}$   $\left(\because 1 \stackrel{0}{\mathrm{A}}={10}^{-10} \mathrm{m}\right)$

The total energy of electron is given by

$E_n=\frac{-m{Z}^2{e}^4}{8{\epsilon }_0^2{h}^2}\left(\frac{1}{n^2}\right)$

$\Rightarrow E_n\propto m$

For electron in first orbit of hydrogen atom,

$E_1=k{m}_e$                       …(iv)

$\text{ where, }k=\frac{e^4}{8{\epsilon }_0^2{h}^2}=\text{ constant. }$

For muon in first orbit,

$E_1^{\text{'}}=k{m_{\mu }}^{\text{'}}$ 

$=k\times 186m_e$       [from Eq. (ii)]

$=186{ \mathrm{km}}_e$

$=186E_1$             [from Eq. (iv)]

$=186(-13.6)\mathrm{eV}$   (given, E₁ = $-$13.6 eV)

= $-$ 2529.6 eV = $-$ 2.5 keV

$\therefore$These values are closest to that of option (c).