The de-Broglie wavelength L associated with an elementary particle of linear momentum p is bets represented by the graph

Options

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B.
C.
D.

Answer

To solve the problem regarding the de Broglie wavelength \( L \) associated with an elementary particle of linear momentum \( p \), we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is the Planck constant and \( p \) is the linear momentum of the particle.

### Step 2: Identify the relationship between wavelength and momentum
From the formula, we can see that the wavelength \( \lambda \) is inversely proportional to the momentum \( p \):
\[ \lambda \propto \frac{1}{p} \]
This means that as the momentum \( p \) increases, the wavelength \( \lambda \) decreases, and vice versa.

### Step 3: Rewrite the relationship using \( L \)
In this problem, the wavelength is represented by \( L \):
\[ L \propto \frac{1}{p} \]

### Step 4: Analyze the implications of the relationship
- If \( p \) increases, \( L \) decreases.
- If \( p \) decreases, \( L \) increases.

### Step 5: Determine the correct graph representation
Given the relationship \( L \propto \frac{1}{p} \), the graph of \( L \) versus \( p \) should show a hyperbolic decline. This means that the graph will not be a straight line, nor will it be constant or exponential.

### Step 6: Evaluate the options provided
1. **Constant graph**: This is incorrect because \( L \) changes with \( p \).
2. **Straight line graph**: This is incorrect because \( L \) is inversely proportional to \( p \), not directly proportional.
3. **Exponential graph**: This is incorrect as well since the relationship is not exponential.
4. **Hyperbolic graph**: This is the correct representation as it shows the inverse relationship.

### Conclusion
The best representation of the de Broglie wavelength \( L \) associated with an elementary particle of linear momentum \( p \) is a hyperbolic graph.

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