Which of these particles having the same kinetic energy has the largest de Broglie wavelength ?

Options

A. Electron
B. Alpha particle
C. Proton
D. Neutron

Explanation

To solve the question of which particle has the largest de Broglie wavelength when they all have the same kinetic energy, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( K \) is the kinetic energy.

2. **Identify Constants**:
In this scenario, we know that:
- \( h \) (Planck's constant) is a constant.
- \( K \) (kinetic energy) is the same for all the particles.

3. **Analyze the Relationship**:
From the formula, we can see that the de Broglie wavelength is inversely proportional to the square root of the mass:
\[ \lambda \propto \frac{1}{\sqrt{m}} \]
This means that as the mass (\( m \)) decreases, the wavelength (\( \lambda \)) increases.

4. **Compare the Masses of the Given Particles**:
We need to compare the masses of the four particles:
- **Electron**: \( m_e = 9.1 \times 10^{-31} \) kg
- **Proton**: \( m_p = 1.67 \times 10^{-27} \) kg
- **Neutron**: \( m_n = 1.67 \times 10^{-27} \) kg
- **Alpha Particle**: \( m_{\alpha} \approx 4 \times m_p \approx 6.68 \times 10^{-27} \) kg

5. **Determine the Particle with the Least Mass**:
Among the particles listed, the electron has the smallest mass. Since the de Broglie wavelength is inversely proportional to the square root of the mass, the electron will have the largest de Broglie wavelength.

6. **Conclusion**:
Therefore, the particle with the largest de Broglie wavelength when all have the same kinetic energy is the **electron**.

### Final Answer:
The electron has the largest de Broglie wavelength among the given particles.

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