The energy that should be added to an electron to reduce its de Broglie wavelength from `lamda " to " lamda/2` is (Considering E to the energy of electron when its de Broglie wavelength is `lamda` )

Options

A. E
B. 2E
C. 3E
D. 4E

Explanation

To solve the problem of finding the energy that should be added to an electron to reduce its de Broglie wavelength from \( \lambda \) to \( \frac{\lambda}{2} \), we can follow these steps:

### Step 1: Understand the relationship between de Broglie wavelength and kinetic energy
The de Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) can also be expressed in terms of kinetic energy \( E \):
\[ p = \sqrt{2mE} \]
where \( m \) is the mass of the electron.

### Step 2: Substitute the expression for 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 3: Relate the initial and final wavelengths with their corresponding energies
When the de Broglie wavelength changes from \( \lambda \) to \( \frac{\lambda}{2} \), we can set up the following relationship:
\[ \frac{\lambda}{\frac{\lambda}{2}} = \frac{\sqrt{2mE'}}{\sqrt{2mE}} \]
where \( E \) is the initial kinetic energy and \( E' \) is the final kinetic energy.

### Step 4: Simplify the equation
This simplifies to:
\[ 2 = \sqrt{\frac{E'}{E}} \]
Squaring both sides gives:
\[ 4 = \frac{E'}{E} \]
Thus, we find:
\[ E' = 4E \]

### Step 5: Calculate the energy that needs to be added
The energy that needs to be added to the electron is given by:
\[ \text{Energy added} = E' - E = 4E - E = 3E \]

### Conclusion
The energy that should be added to the electron to reduce its de Broglie wavelength from \( \lambda \) to \( \frac{\lambda}{2} \) is:
\[ \boxed{3E} \]