An electron, a proton and an alpha particle have kinetic energy of `16E,4E` and `E` respectively. What is the qualitavtive order of their de Broglie wavelengths :-

Options

A. `lambda_(e)gtlambda_(p)=lambda_(alpha)`
B. `lambda_(p)=lambda_(alpha)gtlambda_(e)`
C. `lambda_(p)gtlambda_(e)gtlambda_(alpha)`
D. `lambda_(alpha)ltlambda_(e)gtgtlambda_(alpha)`

Explanation

To determine the qualitative order of the de Broglie wavelengths of an electron, a proton, and an alpha particle with given kinetic energies, 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 \( p \) is the momentum of the particle. For a particle with kinetic energy \( KE \), the momentum can be expressed as:
\[ p = \sqrt{2m \cdot KE} \]
Thus, substituting for momentum in the de Broglie wavelength formula, we have:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

### Step 2: Write the expression for each particle
Given the kinetic energies:
- For the electron: \( KE_e = 16E \)
- For the proton: \( KE_p = 4E \)
- For the alpha particle: \( KE_{\alpha} = E \)

We can write the de Broglie wavelengths for each particle:

1. **Electron**:
\[ \lambda_e = \frac{h}{\sqrt{2m_e \cdot 16E}} = \frac{h}{\sqrt{32m_e E}} \]

2. **Proton**:
\[ \lambda_p = \frac{h}{\sqrt{2m_p \cdot 4E}} = \frac{h}{\sqrt{8m_p E}} \]

3. **Alpha Particle** (which has a mass of approximately \( 4m_p \)):
\[ \lambda_{\alpha} = \frac{h}{\sqrt{2(4m_p) \cdot E}} = \frac{h}{\sqrt{8m_p E}} \]

### Step 3: Compare the wavelengths qualitatively
Now we can compare the expressions for the wavelengths:

- For the electron:
\[ \lambda_e = \frac{h}{\sqrt{32m_e E}} \]

- For the proton:
\[ \lambda_p = \frac{h}{\sqrt{8m_p E}} \]

- For the alpha particle:
\[ \lambda_{\alpha} = \frac{h}{\sqrt{8(4m_p) E}} = \frac{h}{\sqrt{32m_p E}} \]

### Step 4: Determine the order
Since \( m_e \) (mass of the electron) is much smaller than \( m_p \) (mass of the proton), we can conclude that:
- \( \lambda_e \) will be the largest because it is inversely proportional to the square root of the smallest mass.
- \( \lambda_p \) and \( \lambda_{\alpha} \) will be smaller, and since \( \lambda_p \) and \( \lambda_{\alpha} \) have the same denominator, we can say:
\[ \lambda_p = \lambda_{\alpha} \]

### Conclusion
The qualitative order of their de Broglie wavelengths is:
\[ \lambda_e > \lambda_p = \lambda_{\alpha} \]