A proton, a neutron, an electron and an `alpha`-particle have same energy. Then their de-Broglie wavelengths compare as

Options

A. `lamda_(p)=lamda_(n)gtlamda_(e)gtlamda_(alpha)`
B. `lamda_(alpha)ltlamda_(p)=lamda_(n)ltlamda_(e)`
C. `lamda_(e)ltlamda_(p)=lamda_(n)gtlamda_(alpha)`
D. `lamda_(e)=lamda_(p)=lamda_(n)=lamda_(alpha)`

Explanation

To compare the de-Broglie wavelengths of a proton, neutron, electron, and alpha particle when they all have the same energy, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de-Broglie Wavelength Formula**:
The de-Broglie wavelength (λ) 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 Energy**:
The momentum \( p \) can be expressed in terms of kinetic energy \( K \):
\[ p = \sqrt{2mK} \]
where \( m \) is the mass of the particle and \( K \) is the kinetic energy.

3. **Substitute Momentum in the Wavelength Formula**:
Substituting the expression for momentum into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]
Since all particles have the same energy \( K \), we can simplify this to:
\[ \lambda \propto \frac{1}{\sqrt{m}} \]

4. **Compare the Masses of the Particles**:
- Mass of an electron \( m_e \)
- Mass of a proton \( m_p \)
- Mass of a neutron \( m_n \) (approximately equal to \( m_p \))
- Mass of an alpha particle \( m_{\alpha} \) (approximately \( 4m_p \))

From this, we know:
\[ m_{\alpha} > m_p \approx m_n > m_e \]

5. **Determine the Wavelengths**:
Since \( \lambda \) is inversely proportional to the square root of the mass, we can conclude:
- The alpha particle has the largest mass, hence the smallest wavelength.
- The proton and neutron have approximately the same mass, so they will have the same wavelength, which is larger than that of the alpha particle.
- The electron has the smallest mass, thus it will have the largest wavelength.

6. **Final Comparison**:
Therefore, the order of the de-Broglie wavelengths from largest to smallest is:
\[ \lambda_e > \lambda_p = \lambda_n > \lambda_{\alpha} \]

### Summary of Results:
- \( \lambda_e > \lambda_p = \lambda_n > \lambda_{\alpha} \)