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.
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D.

Answer

To solve the question 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 Planck's constant and \( p \) is the linear momentum of the particle.

### Step 2: Relate the wavelength to momentum
From the formula, we can express the relationship between the wavelength \( L \) (which is used in the question) and the linear momentum \( p \):
\[ L = \frac{h}{p} \]
This indicates that the wavelength \( L \) is inversely proportional to the linear momentum \( p \).

### Step 3: Analyze the relationship
Since \( L \) is inversely proportional to \( p \), we can write:
\[ L \propto \frac{1}{p} \]
This means that as the linear momentum \( p \) increases, the wavelength \( L \) decreases, and vice versa.

### Step 4: Determine the graph representation
Given the inverse relationship, we can conclude that the graph of \( L \) versus \( p \) will be a hyperbola, which typically has the following characteristics:
- As \( p \) increases, \( L \) decreases.
- As \( p \) approaches zero, \( L \) approaches infinity.

### Step 5: Identify the correct graph
From the options provided in the question, the correct representation of the relationship \( L \propto \frac{1}{p} \) will be a graph that shows a decrease in \( L \) as \( p \) increases, which is characteristic of a hyperbolic curve.

### Conclusion
Thus, the correct option representing the de-Broglie wavelength \( L \) associated with an elementary particle of linear momentum \( p \) is the one that shows an inverse relationship, typically represented by a hyperbola.

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