The increasing order of wavelength for `He^(+)` ion, neutron (n) and electron `(e^-)` particles, moving with the same velocity is

Options

A. `lambda_(He^+) lt lambda_(e ) lt lambda_(n)`
B. `lambda_(He^+) = lambda_(n) = lambda_(e )`
C. `lambda_(He^+) lt lambda_(n) lt lambda_(e )`
D. `lambda_(e ) lt lambda_(n) lt lambda_(He^+)`

Answer

To determine the increasing order of wavelength for the `He^(+)` ion, neutron (n), and electron `(e^-)` particles moving with the same velocity, we can use the de Broglie wavelength formula:

### Step-by-Step Solution:

1. **Understanding the de Broglie Wavelength**:
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.

2. **Identifying the Relationship**:
From the formula, we can see that the wavelength is inversely proportional to the mass of the particle when the velocity is constant:
\[ \lambda \propto \frac{1}{m} \]
This means that as the mass increases, the wavelength decreases.

3. **Comparing the Masses**:
We need to compare the masses of the three particles:
- Mass of an electron \( m_e \) is approximately \( 9.11 \times 10^{-31} \) kg.
- Mass of a neutron \( m_n \) is approximately \( 1.675 \times 10^{-27} \) kg.
- Mass of a helium ion \( He^{+} \) (which is essentially a helium nucleus) is about \( 4 \times m_p \) (where \( m_p \) is the mass of a proton, approximately \( 1.67 \times 10^{-27} \) kg). Thus, the mass of \( He^{+} \) is around \( 6.64 \times 10^{-27} \) kg.

4. **Ordering the Masses**:
From the above information, we can order the masses from smallest to largest:
- Electron \( (m_e) < Neutron (m_n) < Helium Ion (He^{+}) \)

5. **Determining the Wavelengths**:
Since the wavelength is inversely proportional to mass:
- The electron, having the smallest mass, will have the largest wavelength.
- The neutron will have a larger mass than the electron, so it will have a smaller wavelength than the electron but larger than that of the helium ion.
- The helium ion, having the largest mass, will have the smallest wavelength.

6. **Conclusion**:
Therefore, the increasing order of wavelength is:
\[ \lambda_{He^{+}} < \lambda_{n} < \lambda_{e^{-}} \]
or in terms of the particles:
\[ He^{+} < n < e^{-} \]

### Final Answer:
The increasing order of wavelength for `He^(+)`, neutron (n), and electron `(e^-)` is:
\[ He^{+} < n < e^{-} \]