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 eV. If the…

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 $m^{\prime}$m′ that revolves around the position of the heavier particle. Then, the reduced mass of electron is $$m^{\prime}=\frac{m M}{m+M}$$m′=m+MmM​ where, $m=$m= mass of electron and $$M=\text { mass of nucleus }$$M= mass of nucleus  Its values is less than $m$m. Given, radius of first orbit for electron $\mathrm{r}_1=0.51 Å$r1​=0.51A˚, ground state energy of electron, $\mathrm{E}_1=-13.6 \mathrm{eV}$E1​=−13.6eV, mass of electron $=\mathrm{m}_{\mathrm{e}}$=me​ mass of muon, $\mathrm{m}_\mu=207 \mathrm{~m}_{\mathrm{e}}$mμ​=207 me​ and mass of nucleus, $M=1836 \mathrm{~m}_{\mathrm{e}}$M=1836 me​ When electron in hydrogen atom is replaced by muon, the reduced mass of muon is $$\mathrm{m}_\mu^{\prime}=\frac{\mathrm{m}_\mu \mathrm{M}}{\mathrm{m}_\mu+\mathrm{M}}$$mμ′​=mμ​+Mmμ​M​ Substituting the given values in Eq. (i), we get $$\mathrm{m}_\mu^{\prime}=\frac{207 \mathrm{~m}_{\mathrm{e}} \times 1836 \mathrm{~m}_{\mathrm{e}}}{207 \mathrm{~m}_{\mathrm{e}}+1836 \mathrm{~m}_{\mathrm{e}}} \approx 186 \mathrm{~m}_{\mathrm{e}}$$mμ′​=207 me​+1836 me​207 me​×1836 me​​≈186 me​ The radius of first orbit in hydrogen atom for electron is given by, $$\mathrm{r}_1=\frac{\mathrm{h}^2 \varepsilon_0}{\pi \mathrm{m}_{\mathrm{e}} \mathrm{e}^2}$$r1​=πme​e2h2ε0​​ The radius of first orbit for muon is $$\mathrm{r}_1^{\prime}=\frac{\mathrm{h}^2 \varepsilon_0}{\pi \mathrm{m}_\mu^{\prime} \mathrm{e}^2}$$r1′​=πmμ′​e2h2ε0​​ $$\begin{aligned} & {\left[\because \text { charge of } \mu=\text { charge of } \mathrm{e}^{-}\right]} \\ & =\frac{\mathrm{h}^2 \varepsilon_0}{\pi \times 186 \mathrm{~m}_{\mathrm{e}} \mathrm{e}^2} \\ & =\left(\frac{\mathrm{h}^2 \varepsilon_0}{\pi \mathrm{m}_{\mathrm{e}} \mathrm{e}^2}\right) \frac{\mathrm{l}}{186}=\frac{\mathrm{r}_1}{186} \\ & =\frac{0.51}{186} Å \quad\left[\because \mathrm{r}_1=0.51 Å\right] \\ & =2.74 \times 10^{-13} \mathrm{~m} \quad\left[\because 1 Å=10^{-10} \mathrm{~m}\right] \\ & \end{aligned}$$​[∵ charge of μ= charge of e−]=π×186 me​e2h2ε0​​=(πme​e2h2ε0​​)186l​=186r1​​=1860.51​A˚[∵r1​=0.51A˚]=2.74×10−13 m[∵1A˚=10−10 m]​ The total energy of electron is given by $$\mathrm{E}_{\mathrm{n}}=\frac{-\mathrm{mZ}^2 \mathrm{e}^4}{8 \varepsilon_0^2 \mathrm{~h}^2}\left(\frac{1}{\mathrm{n}^2}\right) \Rightarrow \mathrm{E}_{\mathrm{n}} \propto \mathrm{m}$$En​=8ε02​ h2−mZ2e4​(n21​)⇒En​∝m For electron in first orbit of hydrogen atom, $$\mathrm{E}_1=\mathrm{km}_{\mathrm{e}}$$E1​=kme​ where, $\mathrm{k}=\frac{\mathrm{e}^4}{8 \varepsilon_0^2 \mathrm{~h}^2}=$k=8ε02​ h2e4​= constant. For muon in first orbit $$\begin{array}{rlr} \mathrm{E}_1{ }^{\prime} & =\mathrm{km}_\mu{ }^{\prime} & \\ & =\mathrm{k} \times 186 \mathrm{~m}_{\mathrm{e}} & \text { [from Eq. (i)] } \\ & =186 \mathrm{~km}_{\mathrm{e}} & \\ & =186 \mathrm{E}_1 & \text { [from Eq. (iv)] } \\ & =186(-13.6) \mathrm{eV} & \text { (given) } \\ & =-2529.6 \mathrm{eV} & \\ & =-2.5 \mathrm{keV} & \end{array}$$E1​′​=kmμ​′=k×186 me​=186 kme​=186E1​=186(−13.6)eV=−2529.6eV=−2.5keV​ [from Eq. (i)]  [from Eq. (iv)]  (given) ​ $\therefore$∴ The values are closest to that of options (3).