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 particles `He^(+)`, neutron (n), and electron `(e^-)`, we will use the de Broglie wavelength formula:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
Where:
- \( \lambda \) = wavelength
- \( h \) = Planck's constant (approximately \( 6.626 \times 10^{-34} \, \text{Js} \))
- \( m \) = mass of the particle
- \( v \) = velocity of the particle

### Step 2: Analyze 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.

### Step 3: Compare the masses of the particles
Now, we need to compare the masses of the three particles:
- Mass of `He^(+)` (Helium ion) = \( 6.7 \times 10^{-27} \, \text{kg} \)
- Mass of neutron (n) = \( 1.67 \times 10^{-27} \, \text{kg} \)
- Mass of electron (e) = \( 9.1 \times 10^{-31} \, \text{kg} \)

### Step 4: Order the masses
Now we can order the masses from largest to smallest:
1. Mass of `He^(+)` > Mass of neutron > Mass of electron
\[ 6.7 \times 10^{-27} \, \text{kg} > 1.67 \times 10^{-27} \, \text{kg} > 9.1 \times 10^{-31} \, \text{kg} \]

### Step 5: Determine the order of wavelengths
Since the wavelength is inversely proportional to mass, we can conclude:
- The particle with the largest mass (`He^(+)`) will have the smallest wavelength.
- The particle with the medium mass (neutron) will have a medium wavelength.
- The particle with the smallest mass (electron) will have the largest wavelength.

Thus, the order of wavelengths from smallest to largest is:
\[ \lambda_{He^{+}} < \lambda_{n} < \lambda_{e^{-}} \]

### Final Answer
The increasing order of wavelength is:
\[ \lambda_{He^{+}} < \lambda_{n} < \lambda_{e^{-}} \]