In a hydrogen atom following the Bohr's psotulates the product of linear momentum and angular momentum is proportional ot `(n)^(x)` where 'n' is the orbit number. Find the valur of x.

Answer

To solve the problem, we need to find the value of \( x \) such that the product of linear momentum \( P \) and angular momentum \( L \) in a hydrogen atom is proportional to \( n^x \), where \( n \) is the orbit number.

### Step-by-Step Solution:

1. **Understanding Linear Momentum**:
- The linear momentum \( P \) of an electron in orbit is given by:
\[ P = mv \]
- Here, \( m \) is the mass of the electron and \( v \) is its velocity.

2. **Velocity and Orbit Number**:
- According to Bohr's model, the velocity \( v \) of the electron is inversely proportional to the orbit number \( n \):
\[ v \propto \frac{1}{n} \]
- Therefore, we can express the linear momentum as:
\[ P \propto m \cdot v \propto m \cdot \frac{1}{n} \]
- This implies:
\[ P \propto \frac{1}{n} \]

3. **Understanding Angular Momentum**:
- The angular momentum \( L \) is given by:
\[ L = mvr \]
- According to Bohr's postulate, the angular momentum is quantized and given by:
\[ L = \frac{nh}{2\pi} \]
- This shows that:
\[ L \propto n \]

4. **Product of Linear and Angular Momentum**:
- Now, we need to find the product \( P \cdot L \):
\[ P \cdot L \propto \left(\frac{1}{n}\right) \cdot n \]
- Simplifying this gives:
\[ P \cdot L \propto 1 \]

5. **Expressing in Terms of \( n^x \)**:
- Since \( P \cdot L \) is proportional to a constant (which can be written as \( n^0 \)), we can write:
\[ P \cdot L \propto n^0 \]
- Therefore, we can conclude that:
\[ x = 0 \]

### Final Answer:
The value of \( x \) is \( 0 \).

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