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

Options

A. `lambda_(p)=lambda_(n) gt lambda_(e) gt lambda_(alpha)`
B. `lambda_(alpha) lt lambda_(p)=lambda_(n) lt lambda_(e)`
C. `lambda_(e) lt lambda_(p)=lambda_(n) gt lambda_(alpha)`
D. `lambda_(e) = lambda_(p) = lambda_(n) = lambda_(alpha)`

Explanation

To solve the problem of comparing the de-Broglie wavelengths of a proton, a neutron, an electron, and an alpha particle with the same kinetic energy, we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (\( \lambda \)) 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.

### Step 2: Relate momentum to kinetic energy
The kinetic energy (\( KE \)) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
We can express momentum (\( p \)) in terms of kinetic energy:
\[ p = mv = \sqrt{2m \cdot KE} \]

### Step 3: Substitute momentum into the de-Broglie wavelength formula
Substituting the expression for momentum into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

### Step 4: Analyze the relationship between wavelength and mass
Since the kinetic energy is the same for all particles, 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 increases, the wavelength decreases.

### Step 5: Compare the masses of the particles
- Mass of an electron (\( m_e \)) is the least.
- Mass of a proton (\( m_p \)) is approximately equal to the mass of a neutron (\( m_n \)).
- Mass of an alpha particle (\( m_{\alpha} \)) is the greatest.

### Step 6: Arrange the wavelengths based on mass
Since \( \lambda \) is inversely proportional to \( \sqrt{m} \):
- The electron has the smallest mass, so it has the largest wavelength.
- The proton and neutron have approximately equal masses, so they will have approximately equal wavelengths.
- The alpha particle has the largest mass, so it has the smallest wavelength.

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

### Conclusion
The final comparison of the de-Broglie wavelengths is:
\[ \lambda_{\alpha} < \lambda_p = \lambda_n < \lambda_e \]