Which of the following graphs shows the variation of de-Broglie wavelength with potential through which a particle of charge q and mass m is accelerated ?

Options

A.
B.
C.
D.

Answer

To solve the problem of determining the variation of de-Broglie wavelength (\(\lambda\)) with the potential (\(V\)) through which a charged particle (with charge \(q\) and mass \(m\)) is accelerated, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential
When a particle with charge \(q\) is accelerated through a potential difference \(V\), its kinetic energy (\(KE\)) can be expressed as:
\[ KE = qV \]

### Step 2: Relate kinetic energy to momentum
The kinetic energy can also be expressed in terms of momentum (\(p\)):
\[ KE = \frac{p^2}{2m} \]
Equating the two expressions for kinetic energy gives:
\[ qV = \frac{p^2}{2m} \]

### Step 3: Solve for momentum
Rearranging the equation to find momentum \(p\):
\[ p^2 = 2mqV \]
\[ p = \sqrt{2mqV} \]

### Step 4: Use the de-Broglie wavelength formula
The de-Broglie wavelength (\(\lambda\)) is given by:
\[ \lambda = \frac{h}{p} \]
Substituting the expression for momentum we derived:
\[ \lambda = \frac{h}{\sqrt{2mqV}} \]

### Step 5: Analyze the relationship between \(\lambda\) and \(V\)
From the equation \(\lambda = \frac{h}{\sqrt{2mqV}}\), we can see that:
\[ \lambda \propto \frac{1}{\sqrt{V}} \]
This indicates that as the potential \(V\) increases, the de-Broglie wavelength \(\lambda\) decreases.

### Step 6: Determine the nature of the graph
Since \(\lambda\) is inversely proportional to the square root of \(V\), we can conclude that the graph of \(\lambda\) versus \(V\) will not be a straight line. Instead, it will be a curve that decreases as \(V\) increases.

### Step 7: Identify the correct graph
From the analysis, we can infer that:
- As \(V\) approaches 0, \(\lambda\) approaches infinity.
- As \(V\) increases, \(\lambda\) decreases towards 0.

This behavior corresponds to a graph that starts high (when \(V\) is low) and decreases as \(V\) increases, which is characteristic of a hyperbolic or inverse relationship.

### Conclusion
Based on the analysis, the correct graph that shows the variation of de-Broglie wavelength with potential is **Option 2**.

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