An electron, a proton and an alpha particle have KE of 16E, 4E and E respectively. What is the qualitative order of their de-Broglie wavelengths?

Options

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

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 will follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2mK_e}} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( K_e \) is the kinetic energy of the particle.

### Step 2: Identify the kinetic energies
The kinetic energies given are:
- For the electron: \( K_e = 16E \)
- For the proton: \( K_p = 4E \)
- For the alpha particle: \( K_{\alpha} = E \)

### Step 3: Compare the masses of the particles
- Mass of the electron (\( m_e \)): \( 9.1 \times 10^{-31} \) kg (very small)
- Mass of the proton (\( m_p \)): approximately \( 1.67 \times 10^{-27} \) kg (much larger than the electron)
- Mass of the alpha particle (\( m_{\alpha} \)): approximately \( 4 \times m_p \) (even larger than the proton)

### Step 4: Calculate the qualitative order of de-Broglie wavelengths
Using the formula for de-Broglie wavelength, we can see that:
- For the electron:
\[ \lambda_e \propto \frac{1}{\sqrt{m_e \cdot 16E}} \quad \text{(mass is very small)} \]
- For the proton:
\[ \lambda_p \propto \frac{1}{\sqrt{m_p \cdot 4E}} \]
- For the alpha particle:
\[ \lambda_{\alpha} \propto \frac{1}{\sqrt{4m_p \cdot E}} \]

### Step 5: Analyze the relationships
1. Since the mass of the electron is much smaller than that of the proton and alpha particle, its wavelength will be larger.
2. The proton and alpha particle have larger masses, and since the alpha particle is four times the mass of the proton, its wavelength will be smaller than that of the proton.

### Conclusion
From the analysis, we can conclude:
- \( \lambda_e > \lambda_p \)
- \( \lambda_p \approx \lambda_{\alpha} \)

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

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