The kinetic energy of an electron, `alpha -particle and a proton are given as 4 K, 2 K and K respectively. The de-Broglie wavelength associated with electron `(lambdae)`, `alpha`-particle `(lambda α) `and the proton `(lambda p) are as follows:

Options

A. `lambda α = lambda p gt lambda e`
B. `lambda α ltlambda p ltlambda e`
C. `lambda α = lamnbda p lt lambda e`
D. `lambda α gt lambda p gt lambda e`

Answer

To solve the problem, we need to find the de-Broglie wavelengths associated with an electron, an alpha particle, and a proton based on their given kinetic energies. The kinetic energies are as follows:

- Kinetic Energy of Electron (KE_e) = 4K
- Kinetic Energy of Alpha Particle (KE_α) = 2K
- Kinetic Energy of Proton (KE_p) = K

### Step 1: Write the formula for de-Broglie wavelength
The de-Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the particle and \( h \) is Planck's constant.

### Step 2: Relate kinetic energy to momentum
The kinetic energy (KE) is related to momentum (p) by the formula:
\[ KE = \frac{p^2}{2m} \]
From this, we can express momentum as:
\[ p = \sqrt{2m \cdot KE} \]

### Step 3: Calculate the de-Broglie wavelength for each particle

#### For the Electron:
1. **Mass of Electron (m_e)**: \( m_e \)
2. **Kinetic Energy (KE_e)**: \( 4K \)
3. **Momentum (p_e)**:
\[ p_e = \sqrt{2m_e \cdot 4K} = \sqrt{8m_e K} \]
4. **De-Broglie Wavelength (λ_e)**:
\[ \lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{8m_e K}} = \frac{h}{2\sqrt{2m_e K}} \]

#### For the Alpha Particle:
1. **Mass of Alpha Particle (m_α)**: \( 4m_p \) (approximately 4 times the mass of a proton)
2. **Kinetic Energy (KE_α)**: \( 2K \)
3. **Momentum (p_α)**:
\[ p_α = \sqrt{2(4m_p) \cdot 2K} = \sqrt{16m_p K} = 4\sqrt{m_p K} \]
4. **De-Broglie Wavelength (λ_α)**:
\[ \lambda_α = \frac{h}{p_α} = \frac{h}{4\sqrt{m_p K}} \]

#### For the Proton:
1. **Mass of Proton (m_p)**: \( m_p \)
2. **Kinetic Energy (KE_p)**: \( K \)
3. **Momentum (p_p)**:
\[ p_p = \sqrt{2m_p \cdot K} = \sqrt{2m_p K} \]
4. **De-Broglie Wavelength (λ_p)**:
\[ \lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2m_p K}} \]

### Step 4: Compare the wavelengths
Now we have:
- \( \lambda_e = \frac{h}{2\sqrt{2m_e K}} \)
- \( \lambda_α = \frac{h}{4\sqrt{m_p K}} \)
- \( \lambda_p = \frac{h}{\sqrt{2m_p K}} \)

To compare these wavelengths, we can analyze the ratios:
1. **Comparing \( \lambda_e \) and \( \lambda_p \)**:
\[ \lambda_e = \frac{h}{2\sqrt{2m_e K}} \quad \text{and} \quad \lambda_p = \frac{h}{\sqrt{2m_p K}} \]
Since \( m_e \) is much smaller than \( m_p \), \( \lambda_e \) will be greater than \( \lambda_p \).

2. **Comparing \( \lambda_α \) and \( \lambda_p \)**:
\[ \lambda_α = \frac{h}{4\sqrt{m_p K}} \quad \text{and} \quad \lambda_p = \frac{h}{\sqrt{2m_p K}} \]
Here, \( \lambda_α \) will be greater than \( \lambda_p \) since \( 4\sqrt{m_p K} \) is larger than \( \sqrt{2m_p K} \).

3. **Final Comparison**:
Thus, we can conclude:
\[ \lambda_α > \lambda_p > \lambda_e \]

### Final Answer:
The order of the de-Broglie wavelengths is:
\[ \lambda_α > \lambda_p > \lambda_e \]