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

Key Idea Hydrogen atom can be considered to be the system of two charges, positive charged nucleus and negative charged electron. A system of this kind is equivalent to a single particle of mass $m^{\prime}$ that revolves around the position of the heavier particle. Then, the reduced mass of electron is $$ m^{\prime}=\frac{m M}{m+M} $$ where, $m=$ mass of electron and $$ M=\text { mass of nucleus } $$ Its values is less than $m$. Given, radius of first orbit for electron $\mathrm{r}_1=0.51 Å$, ground state energy of electron, $\mathrm{E}_1=-13.6 \mathrm{eV}$, mass of electron $=\mathrm{m}_{\mathrm{e}}$ mass of muon, $\mathrm{m}_\mu=207 \mathrm{~m}_{\mathrm{e}}$ and mass of nucleus, $M=1836 \mathrm{~m}_{\mathrm{e}}$ When electron in hydrogen atom is replaced by muon, the reduced mass of muon is $$ \mathrm{m}_\mu^{\prime}=\frac{\mathrm{m}_\mu \mathrm{M}}{\mathrm{m}_\mu+\mathrm{M}} $$ Substituting the given values in Eq. (i), we get $$ \mathrm{m}_\mu^{\prime}=\frac{207 \mathrm{~m}_{\mathrm{e}} \times 1836 \mathrm{~m}_{\mathrm{e}}}{207 \mathrm{~m}_{\mathrm{e}}+1836 \mathrm{~m}_{\mathrm{e}}} \approx 186 \mathrm{~m}_{\mathrm{e}} $$ The radius of first orbit in hydrogen atom for electron is given by, $$ \mathrm{r}_1=\frac{\mathrm{h}^2 \varepsilon_0}{\pi \mathrm{m}_{\mathrm{e}} \mathrm{e}^2} $$ The radius of first orbit for muon is $$ \mathrm{r}_1^{\prime}=\frac{\mathrm{h}^2 \varepsilon_0}{\pi \mathrm{m}_\mu^{\prime} \mathrm{e}^2} $$ $$ \begin{aligned} & {\left[\because \text { charge of } \mu=\text { charge of } \mathrm{e}^{-}\right]} \\ & =\frac{\mathrm{h}^2 \varepsilon_0}{\pi \times 186 \mathrm{~m}_{\mathrm{e}} \mathrm{e}^2} \\ & =\left(\frac{\mathrm{h}^2 \varepsilon_0}{\pi \mathrm{m}_{\mathrm{e}} \mathrm{e}^2}\right) \frac{\mathrm{l}}{186}=\frac{\mathrm{r}_1}{186} \\ & =\frac{0.51}{186} Å \quad\left[\because \mathrm{r}_1=0.51 Å\right] \\ & =2.74 \times 10^{-13} \mathrm{~m} \quad\left[\because 1 Å=10^{-10} \mathrm{~m}\right] \\ & \end{aligned} $$ The total energy of electron is given by $$ \mathrm{E}_{\mathrm{n}}=\frac{-\mathrm{mZ}^2 \mathrm{e}^4}{8 \varepsilon_0^2 \mathrm{~h}^2}\left(\frac{1}{\mathrm{n}^2}\right) \Rightarrow \mathrm{E}_{\mathrm{n}} \propto \mathrm{m} $$ For electron in first orbit of hydrogen atom, $$ \mathrm{E}_1=\mathrm{km}_{\mathrm{e}} $$ where, $\mathrm{k}=\frac{\mathrm{e}^4}{8 \varepsilon_0^2 \mathrm{~h}^2}=$ constant. For muon in first orbit $$ \begin{array}{rlr} \mathrm{E}_1{ }^{\prime} & =\mathrm{km}_\mu{ }^{\prime} & \\ & =\mathrm{k} \times 186 \mathrm{~m}_{\mathrm{e}} & \text { [from Eq. (i)] } \\ & =186 \mathrm{~km}_{\mathrm{e}} & \\ & =186 \mathrm{E}_1 & \text { [from Eq. (iv)] } \\ & =186(-13.6) \mathrm{eV} & \text { (given) } \\ & =-2529.6 \mathrm{eV} & \\ & =-2.5 \mathrm{keV} & \end{array} $$ $\therefore$ The values are closest to that of options (3).