Assuming the velocity to be same , which sub-atomic particle possesses smallest de-Broglie wavelength :

Options

A. An electron
B. A proton
C. An `alpha`-particle
D. All will have same value of wavelength

Explanation

To solve the question of which sub-atomic particle possesses the smallest de-Broglie wavelength, 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 (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \(p\) is the momentum of the particle.

### Step 2: Relate momentum to mass and velocity
Momentum (\(p\)) can be expressed as:
\[ p = mv \]
where:
- \(m\) is the mass of the particle,
- \(v\) is its velocity.

### Step 3: Substitute momentum into the de-Broglie wavelength formula
Substituting the expression for momentum into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{mv} \]

### Step 4: Analyze the relationship between wavelength and mass
From the equation \(\lambda = \frac{h}{mv}\), we can see that if the velocity (\(v\)) is constant, then:
\[ \lambda \propto \frac{1}{m} \]
This means that the wavelength is inversely proportional to the mass of the particle. Therefore, a particle with a larger mass will have a smaller de-Broglie wavelength.

### Step 5: Compare the masses of the sub-atomic particles
The sub-atomic particles we need to compare are:
- Electron
- Proton
- Alpha particle

The approximate masses are:
- Mass of electron (\(m_e\)) ≈ \(9.11 \times 10^{-31} \, \text{kg}\)
- Mass of proton (\(m_p\)) ≈ \(1.67 \times 10^{-27} \, \text{kg}\)
- Mass of alpha particle (\(m_{\alpha}\)) ≈ \(4 \times m_p \approx 6.68 \times 10^{-27} \, \text{kg}\)

### Step 6: Determine which particle has the largest mass
From the above values, we can conclude:
- The electron has the smallest mass.
- The proton has a larger mass than the electron.
- The alpha particle has the largest mass of the three.

### Step 7: Identify the particle with the smallest de-Broglie wavelength
Since the de-Broglie wavelength is inversely proportional to mass, the particle with the largest mass (the alpha particle) will have the smallest de-Broglie wavelength.

### Conclusion
Thus, the sub-atomic particle that possesses the smallest de-Broglie wavelength is the **alpha particle**.

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