Electron, proton and `He^(++)` are moving with same K.E.Then order of de-broglie wavelengths are:

Options

A. `lambda_e lt lambda_p lt lambda_(He^(++))`
B. `lambda_p gt lambda_e gt lambda_(He^(++))`
C. `lambda_e gt lambda_(He^(++)) gt lambda_p`
D. `lambda_e gt lambda_p gt lambda_(He^(++))`

Answer

To solve the problem of finding the order of de Broglie wavelengths for an electron, proton, and `He^(++)` (helium nucleus), we start by using the relationship between de Broglie wavelength and momentum.

### Step-by-Step Solution:

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.

2. **Relate Momentum to Kinetic Energy**:
The kinetic energy (K.E.) of a particle is related to its momentum by the equation:
\[ K.E. = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot K.E.} \]
where \( m \) is the mass of the particle.

3. **Substitute Momentum into the Wavelength Formula**:
Substituting the expression for momentum into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2m \cdot K.E.}} \]

4. **Analyze the Relationship**:
Since the kinetic energy is the same for all three particles, we can see that the de Broglie wavelength is inversely proportional to the square root of the mass:
\[ \lambda \propto \frac{1}{\sqrt{m}} \]

5. **Identify the Masses**:
- Mass of electron (\( m_e \)): \( 9.1 \times 10^{-31} \, kg \)
- Mass of proton (\( m_p \)): \( 1.67 \times 10^{-27} \, kg \)
- Mass of helium nucleus (\( He^{++} \)): Since helium has an atomic mass of approximately 4, the mass is \( 4 \times 1.67 \times 10^{-27} \, kg = 6.68 \times 10^{-27} \, kg \)

6. **Calculate the Ratios of Wavelengths**:
The ratio of the de Broglie wavelengths can be expressed as:
\[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}, \quad \frac{\lambda_e}{\lambda_{He}} = \sqrt{\frac{m_{He}}{m_e}}, \quad \frac{\lambda_p}{\lambda_{He}} = \sqrt{\frac{m_{He}}{m_p}} \]

7. **Calculate Each Ratio**:
- For electron and proton:
\[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{1.67 \times 10^{-27}}{9.1 \times 10^{-31}}} \approx \sqrt{1836} \approx 42.9 \]
- For electron and helium nucleus:
\[ \frac{\lambda_e}{\lambda_{He}} = \sqrt{\frac{6.68 \times 10^{-27}}{9.1 \times 10^{-31}}} \approx \sqrt{73.3} \approx 8.57 \]
- For proton and helium nucleus:
\[ \frac{\lambda_p}{\lambda_{He}} = \sqrt{\frac{6.68 \times 10^{-27}}{1.67 \times 10^{-27}}} \approx \sqrt{4} = 2 \]

8. **Order of Wavelengths**:
From the ratios calculated, we can conclude:
\[ \lambda_e > \lambda_p > \lambda_{He} \]
Thus, the order of de Broglie wavelengths is:
\[ \lambda_e : \lambda_p : \lambda_{He} = 1 : \frac{1}{42.9} : \frac{1}{8.57} \]

### Final Conclusion:
The order of de Broglie wavelengths is:
\[ \text{Electron} > \text{Proton} > \text{Helium nucleus} \]