A muon is a particle that has the same charge as that of an electron but is 200 times revolves round a proton instead of an electron then what will be the orbital magnetic moment of the muon in the ground state of such an atom ? [ Given that Bohr magneton =`9.28 xx10^(24) Am^(2)`]

Options

A. independent of which orbit it is n
B. negative
C. postive
D. decreases with the quantum number n

Explanation

To find the orbital magnetic moment of a muon revolving around a proton in the ground state of such an atom, we can follow these steps:

### Step 1: Understand the properties of the muon
A muon has the same charge as an electron but is approximately 200 times heavier. The charge of the muon is denoted as \( e \), and its mass is \( m_\mu = 200 m_e \), where \( m_e \) is the mass of the electron.

### Step 2: Recall the formula for orbital magnetic moment
The orbital magnetic moment \( \mu \) of a charged particle moving in a circular orbit can be expressed as:
\[ \mu = \frac{e L}{2m} \]
where \( L \) is the angular momentum of the particle, \( e \) is the charge, and \( m \) is the mass of the particle.

### Step 3: Determine the angular momentum in the ground state
In quantum mechanics, the angular momentum \( L \) of an electron in the ground state (n=1) of a hydrogen atom is given by:
\[ L = n \frac{h}{2\pi} = \frac{h}{2\pi} \]
For the muon, since it is in the same quantum state (n=1), we have:
\[ L = \frac{h}{2\pi} \]

### Step 4: Substitute values into the magnetic moment formula
Substituting the values into the magnetic moment formula:
\[ \mu = \frac{e \left(\frac{h}{2\pi}\right)}{2m_\mu} \]
Substituting \( m_\mu = 200 m_e \):
\[ \mu = \frac{e \left(\frac{h}{2\pi}\right)}{2 \cdot 200 m_e} \]
\[ \mu = \frac{e h}{400 \pi m_e} \]

### Step 5: Relate to Bohr magneton
The Bohr magneton \( \mu_B \) is defined as:
\[ \mu_B = \frac{e h}{4 \pi m_e} \]
Thus, we can express the magnetic moment of the muon in terms of the Bohr magneton:
\[ \mu = \frac{1}{100} \mu_B \]

### Step 6: Calculate the numerical value
Given that the Bohr magneton \( \mu_B = 9.28 \times 10^{-24} \, \text{Am}^2 \):
\[ \mu = \frac{1}{100} \times 9.28 \times 10^{-24} \, \text{Am}^2 = 9.28 \times 10^{-26} \, \text{Am}^2 \]

### Final Answer
The orbital magnetic moment of the muon in the ground state of such an atom is:
\[ \mu = 9.28 \times 10^{-26} \, \text{Am}^2 \]

---