Which of the following products, in a hydrogen atom , are independent of the principal quantum number `n` ? The symbels have their usual meanings.

Options

A. `omega^(2) r`
B. `(E)/(v^(2))`
C. `v^(2) r`
D. `(E)/( c)`

Explanation

To determine which products in a hydrogen atom are independent of the principal quantum number \( n \), we need to analyze the relationships between various quantities in the hydrogen atom.

### Step-by-Step Solution:

1. **Understand the Relationships**:
- The velocity \( v \) of an electron in a hydrogen atom is inversely proportional to the principal quantum number \( n \):
\[ v \propto \frac{1}{n} \]
- The energy \( E \) of the electron is inversely proportional to the square of \( n \):
\[ E \propto \frac{1}{n^2} \]
- The radius \( r \) of the electron's orbit is proportional to the square of \( n \):
\[ r \propto n^2 \]

2. **Evaluate Each Option**:
- **Option 1: \( \omega^2 r \)**:
- We know that \( \omega \) (angular frequency) can be expressed as:
\[ \omega = \frac{v}{r} \]
- Thus, \( \omega^2 = \left(\frac{v}{r}\right)^2 \).
- Substituting the proportionalities:
\[ \omega \propto \frac{1/n}{n^2} = \frac{1}{n^3} \implies \omega^2 \propto \frac{1}{n^6} \]
- Therefore, \( \omega^2 r \) becomes:
\[ \omega^2 r \propto \frac{1}{n^6} \cdot n^2 = \frac{1}{n^4} \]
- This indicates that \( \omega^2 r \) is dependent on \( n \).

- **Option 2: \( \frac{E}{v^2} \)**:
- Substituting the proportionalities:
\[ \frac{E}{v^2} \propto \frac{1/n^2}{(1/n)^2} = \frac{1/n^2}{1/n^2} = 1 \]
- This shows that \( \frac{E}{v^2} \) is independent of \( n \).

- **Option 3: \( v^2 r \)**:
- Using the relationships:
\[ v^2 r \propto \left(\frac{1}{n}\right)^2 \cdot n^2 = \frac{1}{n^2} \cdot n^2 = 1 \]
- This indicates that \( v^2 r \) is also independent of \( n \).

- **Option 4: \( \frac{E}{c} \)**:
- Since \( c \) is a constant, we have:
\[ \frac{E}{c} \propto \frac{1/n^2}{c} \propto \frac{1}{n^2} \]
- This shows that \( \frac{E}{c} \) is dependent on \( n \).

3. **Conclusion**:
- The products that are independent of the principal quantum number \( n \) are:
- \( \frac{E}{v^2} \)
- \( v^2 r \)

### Final Answer:
The correct options are **Option 2** and **Option 3**.