An electron, a neutron and an alpha particle have same kinetic energy and their de-Broglie wavelength are `lambda_e, lambda_n and lambda_(alpha)` respectively. Which statement is correct about their de-Broglie wavelengths?

Options

A. `lambda_(e ) gt lambda_n gt lambda_(alpha)`
B. `lambda_(e ) lt lambda_n gt lambda_(alpha)`
C. `lambda_(e ) lt lambda_n lt lambda_(alpha)`
D. `lambda_(e ) gt lambda_n lt lambda_(alpha)`

Answer

To solve the problem, we need to analyze the de-Broglie wavelengths of an electron, a neutron, and an alpha particle, given that they all have the same kinetic energy.

### Step-by-Step Solution:

1. **Understand the de-Broglie Wavelength Formula**:
The de-Broglie wavelength (λ) of 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. **Relate Momentum to Kinetic Energy**:
The kinetic energy (KE) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass and \( v \) is the velocity. We can express momentum \( p \) as:
\[ p = mv \]
From the kinetic energy formula, we can derive momentum in terms of kinetic energy:
\[ KE = \frac{1}{2} mv^2 \implies mv = \sqrt{2m \cdot KE} \]
Thus,
\[ p = \sqrt{2m \cdot KE} \]

3. **Substitute Momentum into the de-Broglie Wavelength Formula**:
Now substituting \( p \) into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]
Since the kinetic energy is the same for all three particles, we can see that:
\[ \lambda \propto \frac{1}{\sqrt{m}} \]
This means that the de-Broglie wavelength is inversely proportional to the square root of the mass of the particle.

4. **Compare the Masses of the Particles**:
- Mass of the electron (\( m_e \)) is approximately \( 9.11 \times 10^{-31} \) kg.
- Mass of the neutron (\( m_n \)) is approximately \( 1836 \times m_e \).
- Mass of the alpha particle (\( m_{\alpha} \)) is approximately \( 4 \times m_n \) (since it consists of 2 protons and 2 neutrons).

5. **Determine the Order of Wavelengths**:
Since \( \lambda \propto \frac{1}{\sqrt{m}} \):
- The electron has the smallest mass, so it will have the largest wavelength:
\[ \lambda_e > \lambda_n > \lambda_{\alpha} \]

### Conclusion:
The correct statement about their de-Broglie wavelengths is:
\[ \lambda_e > \lambda_n > \lambda_{\alpha} \]