Q136 / 160 · AP EAMCET 2014 · Physics
The radius of the first orbit of hydrogen is $\mathrm{r}_{\mathrm{H}}$, and the energy in the ground state is - $13.6 \mathrm{eV}$. Considering a $\mu^{-}$-particle with a mass $207 m_e$ revolving round a proton as in hydrogen atom, the energy and radius of proton and $\mu^{-}$- combination respectively in the first orbit are (assume nucleus to be stationary)
- Correct answerA$-13.6 \times 207 \mathrm{eV}, \frac{\mathrm{C}^{\mathrm{H}}}{207}$
- B$-207 \times 13.6 \mathrm{eV}, 207 r_{\mathrm{H}}$
- C$-\frac{13.6}{207} \mathrm{eV}, \frac{\mathrm{F}}{207}$
- D$-\frac{13.6}{207} \mathrm{eV}, 207 \mathrm{r}_{\mathrm{H}}$
Solution
The total energy of $n$th orbit
$E_n=-\frac{m e^4}{8 \varepsilon_0^2 h^2} \cdot \frac{1}{n^2}$
Obviously $E_n \propto m$
$\begin{array}{llll} <br />& \therefore & \frac{E_\mu}{E_e} & =\frac{m_\mu}{m_e} \\<br />\Rightarrow & E_\mu & =\frac{m_\mu}{m_e} \times E_e<br />\end{array}$
Ground state energy of a proton in hydrogen atom,
$\begin{aligned}<br />E_\mu & =-13.6 \times \frac{207 m_e}{m_e} \mathrm{eV} \\<br />& =-13.6 \times 207 \mathrm{eV}<br />\end{aligned}$
$\left(\because m_\mu=207 m_e\right.$, where $m_e$ is the mass of electron $)$
We know that
$r=\frac{\varepsilon_0 h^2 n^2}{207 \pi m_{\mathrm{e}} \mathrm{e}^2}$
For ground state $(n-1)$ for proton, we have
$r_\mu=\frac{\varepsilon_0 h^2}{207 \pi m_e \cdot e^2}$
But $\frac{\varepsilon_0 h^2}{\pi m_{\mathrm{e}} \cdot e^2}=$ ground state radius of hydrogen atom
$\begin{aligned}<br />\frac{\varepsilon_0 h^2}{\pi m_e \cdot e^2} & =r_{\mathrm{H}} \\<br />\therefore \quad r_\mu & =\frac{r_{\mathrm{H}}}{207}<br />\end{aligned}$
$E_n=-\frac{m e^4}{8 \varepsilon_0^2 h^2} \cdot \frac{1}{n^2}$
Obviously $E_n \propto m$
$\begin{array}{llll} <br />& \therefore & \frac{E_\mu}{E_e} & =\frac{m_\mu}{m_e} \\<br />\Rightarrow & E_\mu & =\frac{m_\mu}{m_e} \times E_e<br />\end{array}$
Ground state energy of a proton in hydrogen atom,
$\begin{aligned}<br />E_\mu & =-13.6 \times \frac{207 m_e}{m_e} \mathrm{eV} \\<br />& =-13.6 \times 207 \mathrm{eV}<br />\end{aligned}$
$\left(\because m_\mu=207 m_e\right.$, where $m_e$ is the mass of electron $)$
We know that
$r=\frac{\varepsilon_0 h^2 n^2}{207 \pi m_{\mathrm{e}} \mathrm{e}^2}$
For ground state $(n-1)$ for proton, we have
$r_\mu=\frac{\varepsilon_0 h^2}{207 \pi m_e \cdot e^2}$
But $\frac{\varepsilon_0 h^2}{\pi m_{\mathrm{e}} \cdot e^2}=$ ground state radius of hydrogen atom
$\begin{aligned}<br />\frac{\varepsilon_0 h^2}{\pi m_e \cdot e^2} & =r_{\mathrm{H}} \\<br />\therefore \quad r_\mu & =\frac{r_{\mathrm{H}}}{207}<br />\end{aligned}$
Hence, the correct answer is option: 'A'.