Question

The graph which shows the variation of the de Broglie wavelength (λ) of a particle and its associated momentum (p) is

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The de Broglie hypothesis states that every moving particle exhibits wave-like properties.
The wavelength associated with a particle is known as the de Broglie wavelength (\(\lambda\)).
This wavelength is inversely proportional to the magnitude of the particle's momentum (\(p\)).
Key Formula or Approach:
The mathematical expression for the de Broglie wavelength is:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \text{ J}\cdot\text{s}\)).
Step 2: Detailed Explanation:
From the formula \(\lambda = \frac{h}{p}\), we can deduce the following:
1. Planck's constant \(h\) is a fixed value, so \(\lambda \times p = h = \text{constant}\).
2. This equation is of the mathematical form \(xy = c\), which represents a rectangular hyperbola in a Cartesian coordinate system.
3. As the momentum \(p\) increases, the wavelength \(\lambda\) decreases.
4. As the momentum \(p\) approaches zero, the wavelength \(\lambda\) tends to infinity.
5. As the momentum \(p\) tends to infinity, the wavelength \(\lambda\) approaches zero.
Looking at the provided options:
- Graph (1) shows a direct linear relationship (\(\lambda \propto p\)), which is incorrect.
- Graph (2) shows a linear inverse relationship, which is incorrect as the derivative is constant.
- Graph (3) shows a relationship where \(\lambda\) increases with \(p\), which is incorrect.
- Graph (4) correctly depicts a rectangular hyperbola, showing that \(\lambda\) and \(p\) are inversely proportional.
Step 3: Final Answer:
Therefore, the correct graph is the one represented in option (4).