The magnitude of the de-Broglie wavelength `(lambda)` of an electron `(e)`,proton`(p)`,neutron `(n)` and `alpha` particle `(a)` all having the same energy of `Mev`, in the increasing order will follow the sequence:

Options

A. `lambda_(e),lambda_(p), lambda_(n),lambda_(alpha)`
B. `lambda_(e),lambda_(n),lambda_(p),lambda_(alpha)`
C. `lambda_(alpha), lambda_(n),lambda_(p),lambda_(e)`
D. `lambda_(p), lambda_(e),lamda_(alpha), lambda_(n)`

Explanation

To solve the problem of determining the increasing order of the de Broglie wavelengths of an electron, proton, neutron, and alpha particle, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 2: Relate momentum to kinetic energy
The kinetic energy (E) of a particle is related to its momentum (p) and mass (m) by the equation:
\[ E = \frac{p^2}{2m} \]
From this, we can express momentum in terms of kinetic energy:
\[ p = \sqrt{2mE} \]

### Step 3: Substitute momentum into the de Broglie wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

### Step 4: Analyze the relationship between wavelength and mass
From the equation \( \lambda = \frac{h}{\sqrt{2mE}} \), we can see that the de Broglie wavelength is inversely proportional to the square root of the mass:
\[ \lambda \propto \frac{1}{\sqrt{m}} \]
This means that as the mass increases, the wavelength decreases.

### Step 5: Compare the masses of the particles
Now, we need to compare the masses of the four particles:
- Mass of electron (m_e) is very small.
- Mass of proton (m_p) is approximately 1836 times that of the electron.
- Mass of neutron (m_n) is slightly more than that of the proton (about 1839 times the mass of the electron).
- Mass of alpha particle (m_a) is about 4 times the mass of a helium nucleus, which is roughly 4 times the mass of the proton (approximately 7400 times the mass of the electron).

### Step 6: Determine the order of wavelengths
Since the de Broglie wavelength is inversely proportional to the square root of the mass, we can arrange the particles in increasing order of mass:
- Electron (lightest) → Neutron → Proton → Alpha particle (heaviest)

Thus, the order of de Broglie wavelengths from longest to shortest is:
\[ \lambda_e > \lambda_n > \lambda_p > \lambda_a \]

### Step 7: Write the increasing order of wavelengths
In terms of increasing order of wavelength, we have:
\[ \lambda_a < \lambda_p < \lambda_n < \lambda_e \]

### Final Answer
The increasing order of the de Broglie wavelengths of the particles is:
\[ \text{Alpha particle} < \text{Proton} < \text{Neutron} < \text{Electron} \]