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

Options:
A $0.53 \times 10^{-13} \mathrm{~m},-3.6 \mathrm{eV}$
B $25.6 \times 10^{-13} \mathrm{~m},-2.8 \mathrm{eV}$
C $2.56 \times 10^{-13} \mathrm{~m},-2.8 \mathrm{keV}$
D $2.56 \times 10^{-13} \mathrm{~m},-13.6 \mathrm{eV}$

The correct answer is:
$2.56 \times 10^{-13} \mathrm{~m},-2.8 \mathrm{keV}$

Solution:
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).

Topic: Physics, Atomic Physics, NEET, NEET 2019 (Odisha)