Which of following graphs correctly represents the variation of particle momentum with de-Broglie wavelength?

Options

A.
B.
C.
D.

Answer

To solve the question regarding the variation of particle momentum with de-Broglie wavelength, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de-Broglie Hypothesis**:
- According to de-Broglie's hypothesis, every particle has a wave-like nature. The wavelength (λ) associated with a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Rearranging the Equation**:
- From the de-Broglie equation, we can express momentum \( p \) in terms of wavelength \( \lambda \):
\[ p = \frac{h}{\lambda} \]
- This shows that momentum is inversely proportional to the wavelength.

3. **Graphical Representation**:
- Since \( p \) is inversely proportional to \( \lambda \), we can conclude that as the wavelength increases, the momentum decreases. This relationship is characteristic of a hyperbolic graph.
- If we plot momentum \( p \) on the y-axis and wavelength \( \lambda \) on the x-axis, the graph will be a hyperbola that approaches the axes but never touches them.

4. **Analyzing the Options**:
- **Option A**: Graph showing a direct relationship (incorrect, as it should be hyperbolic).
- **Option B**: Another incorrect representation, likely showing a different incorrect relationship.
- **Option C**: This option may represent the relationship between \( \lambda \) and \( \frac{1}{p} \), which would be a straight line (but not what we need).
- **Option D**: This option should correctly represent the inverse relationship between \( p \) and \( \lambda \) as a hyperbola.

5. **Conclusion**:
- The correct graph that represents the variation of particle momentum with de-Broglie wavelength is the one that depicts an inverse relationship, which is a hyperbola.

### Final Answer:
The correct graph is the one that shows the inverse relationship between momentum \( p \) and de-Broglie wavelength \( \lambda \), which is a hyperbola.

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