The radius of the first orbit of hydrogen is `r_H` and the energy in the ground state is `-13.6 eV`.considering a`mu` - particle with the mass 207`m_e` revolving round a porton as in hydrogen atom, the energy and radius of proton and `mu`-combination respectively in the first orbit are(assume nucleus to be stationary)

Options

A. `-13.6xx207eV,(r_(H))/(207)`
B. `-207xx13.6eV,207r_(H)`
C. `(-13.6)/(207)eV,(r_(H))/(207)`
D. `(-13.6)/(207)eV,207r_(H)`

Answer

To solve the problem, we need to determine the radius and energy of a muon (μ) particle revolving around a proton, similar to how an electron revolves around a proton in a hydrogen atom. The mass of the muon is given as 207 times the mass of the electron (m_e).

### Step-by-Step Solution:

1. **Understanding the Radius of the First Orbit:**
The radius of the first orbit in a hydrogen atom is given by the formula:
\[ r_H = \frac{4 \pi \epsilon_0 h^2}{m_e e^2} \]
where \( \epsilon_0 \) is the permittivity of free space, \( h \) is Planck's constant, \( m_e \) is the mass of the electron, and \( e \) is the charge of the electron.

2. **Finding the Radius for the Muon:**
The radius of the first orbit for a muon revolving around a proton can be expressed as:
\[ r_{\mu} = \frac{4 \pi \epsilon_0 h^2}{m_{\mu} e^2} \]
where \( m_{\mu} = 207 m_e \) (mass of the muon).

The ratio of the radii for the muon and hydrogen atom is:
\[ \frac{r_{\mu}}{r_H} = \frac{m_e}{m_{\mu}} = \frac{m_e}{207 m_e} = \frac{1}{207} \]
Therefore, the radius of the muon orbit is:
\[ r_{\mu} = \frac{r_H}{207} \]

3. **Understanding the Energy of the First Orbit:**
The energy of the electron in the ground state of hydrogen is given as:
\[ E_H = -13.6 \text{ eV} \]

4. **Finding the Energy for the Muon:**
The energy of the muon can be expressed as:
\[ E_{\mu} = -\frac{m_{\mu} e^4}{8 \pi \epsilon_0 h^2 n^2} \]
Since the energy is directly proportional to the mass, we have:
\[ \frac{E_{\mu}}{E_H} = \frac{m_{\mu}}{m_e} = 207 \]
Therefore,
\[ E_{\mu} = -207 \times 13.6 \text{ eV} = -2815.2 \text{ eV} \]

5. **Final Results:**
- The radius of the muon orbit is:
\[ r_{\mu} = \frac{r_H}{207} \]
- The energy of the muon in the ground state is:
\[ E_{\mu} = -2815.2 \text{ eV} \]

### Summary:
- Radius of muon orbit: \( r_{\mu} = \frac{r_H}{207} \)
- Energy of muon: \( E_{\mu} = -2815.2 \text{ eV} \)