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 regarding the de-Broglie wavelengths of an electron, a neutron, 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 (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) can also be expressed in terms of mass (m) and velocity (v) as:
\[ p = mv \]

### Step 2: Relate kinetic energy to velocity
The kinetic energy (KE) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
From this, we can express the velocity \( v \) in terms of kinetic energy and mass:
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

### Step 3: Substitute velocity into the de-Broglie wavelength formula
Substituting the expression for velocity into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{mv} = \frac{h}{m \sqrt{\frac{2 \cdot KE}{m}}} = \frac{h}{\sqrt{2 \cdot KE \cdot m}} \]
This shows that the de-Broglie wavelength is inversely proportional to the square root of the mass:
\[ \lambda \propto \frac{1}{\sqrt{m}} \]

### Step 4: Compare the masses of the particles
Now, we need to compare the masses of the electron, neutron, and alpha particle:
- Mass of electron (\( m_e \)) is the smallest.
- Mass of neutron (\( m_n \)) is greater than that of the electron.
- Mass of alpha particle (\( m_{\alpha} \)) is the largest (approximately 4 times the mass of a proton).

### Step 5: Determine the relationship between wavelengths
Since the de-Broglie wavelength is inversely proportional to the square root of the mass:
- The electron, having the smallest mass, will have the largest wavelength (\( \lambda_e \)).
- The neutron will have a smaller wavelength than the electron but larger than that of the alpha particle (\( \lambda_n \)).
- The alpha particle will have the smallest wavelength (\( \lambda_{\alpha} \)).

Thus, we conclude:
\[ \lambda_e > \lambda_n > \lambda_{\alpha} \]

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