The ratio of angular momentum L to the atomic dipole moment `mu_(i)` for hydrogen like atoms and ions is

Options

A. always constant and is equal to the ratio of mass to the charge of electron
B. always constant and is equal to twice the ratio of mass to the charge of electron to the principal quantum number n?
C. Proportional to the principal quantum number n
D. proportional to `(1)/(n^(2))`

Explanation

To find the ratio of angular momentum \( L \) to the atomic dipole moment \( \mu_i \) for hydrogen-like atoms and ions, we will follow these steps:

### Step 1: Understand the Angular Momentum
The angular momentum \( L \) of an electron revolving in a circular orbit around the nucleus is given by the formula:
\[ L = mvr \]
where:
- \( m \) is the mass of the electron,
- \( v \) is the speed of the electron,
- \( r \) is the radius of the orbit.

### Step 2: Calculate the Atomic Dipole Moment
The atomic dipole moment \( \mu_i \) can be calculated using the formula:
\[ \mu_i = i \cdot A \]
where:
- \( i \) is the current,
- \( A \) is the area of the orbit.

### Step 3: Determine the Current \( i \)
The current \( i \) due to the revolving electron can be expressed as:
\[ i = \frac{q}{T} \]
where:
- \( q \) is the charge of the electron (denoted as \( e \)),
- \( T \) is the time period for one complete revolution.

The time period \( T \) can be calculated as:
\[ T = \frac{2\pi r}{v} \]
Substituting this into the expression for current, we get:
\[ i = \frac{e}{T} = \frac{e \cdot v}{2\pi r} \]

### Step 4: Calculate the Area \( A \)
The area \( A \) for a circular orbit is:
\[ A = \pi r^2 \]

### Step 5: Substitute into the Dipole Moment Formula
Now substituting \( i \) and \( A \) into the dipole moment formula:
\[ \mu_i = \left(\frac{e \cdot v}{2\pi r}\right) \cdot (\pi r^2) = \frac{e v r}{2} \]

### Step 6: Formulate the Ratio \( \frac{L}{\mu_i} \)
Now, we can find the ratio of angular momentum to the dipole moment:
\[ \frac{L}{\mu_i} = \frac{mvr}{\frac{e v r}{2}} = \frac{mvr \cdot 2}{e v r} \]
Here, \( v \) and \( r \) cancel out:
\[ \frac{L}{\mu_i} = \frac{2m}{e} \]

### Conclusion
The ratio of angular momentum \( L \) to the atomic dipole moment \( \mu_i \) for hydrogen-like atoms and ions is:
\[ \frac{L}{\mu_i} = \frac{2m}{e} \]