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 given their 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:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( E \) is the kinetic energy of the particle.

### Step 2: Identify the kinetic energies
We have the kinetic energies as follows:
- For the electron: \( E_e = 16E \)
- For the proton: \( E_p = 4E \)
- For the alpha particle: \( E_{\alpha} = E \)

### Step 3: Determine the masses of the particles
- Mass of the proton: \( m_p = m \) (let's denote the mass of the proton as \( m \)).
- Mass of the electron: \( m_e = \frac{m}{1800} \) (the mass of the electron is approximately 1/1800 times that of a proton).
- Mass of the alpha particle: \( m_{\alpha} = 4m \) (the alpha particle consists of 2 protons and 2 neutrons, so its mass is approximately 4 times that of a proton).

### Step 4: Calculate the de Broglie wavelengths
Now we can calculate the de Broglie wavelengths for each particle:

1. **For the electron**:
\[ \lambda_e = \frac{h}{\sqrt{2 \cdot \frac{m}{1800} \cdot 16E}} = \frac{h}{\sqrt{\frac{32mE}{1800}}} = \frac{h \sqrt{1800}}{\sqrt{32mE}} \]

2. **For the proton**:
\[ \lambda_p = \frac{h}{\sqrt{2m \cdot 4E}} = \frac{h}{\sqrt{8mE}} = \frac{h}{\sqrt{8mE}} \]

3. **For the alpha particle**:
\[ \lambda_{\alpha} = \frac{h}{\sqrt{2 \cdot 4m \cdot E}} = \frac{h}{\sqrt{8mE}} = \frac{h}{\sqrt{8mE}} \]

### Step 5: Compare the wavelengths
From the calculations:
- \( \lambda_e \) has a factor of \( \sqrt{1800} \) in the numerator, making it larger.
- \( \lambda_p \) and \( \lambda_{\alpha} \) are equal since they both simplify to \( \frac{h}{\sqrt{8mE}} \).

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

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