Kinetic energy of the particle is E and it's De-Broglie wavelength is `lambda`. On increasing it's KE by `delta E`, it's new De-Broglie wavelength becomes `lambda/2`. Then `delta E` is

Options

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

Explanation

To solve the problem, we need to relate the kinetic energy of a particle to its De-Broglie wavelength and find the increase in kinetic energy, denoted as \(\Delta E\), when the wavelength changes from \(\lambda\) to \(\frac{\lambda}{2}\).

### Step-by-step solution:

1. **Understand the relationship between kinetic energy and De-Broglie wavelength**:
The De-Broglie wavelength \(\lambda\) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

2. **Express momentum in terms of kinetic energy**:
The momentum \(p\) can be expressed in terms of kinetic energy \(E\):
\[ p = \sqrt{2mE} \]
where \(m\) is the mass of the particle. Substituting this into the De-Broglie wavelength equation gives:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

3. **Set up the initial and final conditions**:
Initially, the De-Broglie wavelength is \(\lambda\) corresponding to kinetic energy \(E\). After increasing the kinetic energy by \(\Delta E\), the new kinetic energy becomes \(E + \Delta E\), and the new wavelength is given as:
\[ \lambda' = \frac{h}{\sqrt{2m(E + \Delta E)}} \]
According to the problem, this new wavelength is half of the original wavelength:
\[ \lambda' = \frac{\lambda}{2} \]

4. **Set up the equation**:
From the relation \(\lambda' = \frac{\lambda}{2}\), we can write:
\[ \frac{h}{\sqrt{2m(E + \Delta E)}} = \frac{1}{2} \cdot \frac{h}{\sqrt{2mE}} \]

5. **Cross-multiply and simplify**:
Cross-multiplying gives:
\[ h \cdot \sqrt{2mE} = \frac{1}{2} \cdot h \cdot \sqrt{2m(E + \Delta E)} \]
Canceling \(h\) and simplifying leads to:
\[ \sqrt{2mE} = \frac{1}{2} \sqrt{2m(E + \Delta E)} \]

6. **Square both sides**:
Squaring both sides results in:
\[ 2mE = \frac{1}{4} \cdot 2m(E + \Delta E) \]
Simplifying gives:
\[ 2mE = \frac{1}{2}m(E + \Delta E) \]

7. **Eliminate \(m\) and solve for \(\Delta E\)**:
Dividing through by \(m\) (assuming \(m \neq 0\)):
\[ 2E = \frac{1}{2}(E + \Delta E) \]
Multiplying through by 2 gives:
\[ 4E = E + \Delta E \]
Rearranging this leads to:
\[ \Delta E = 4E - E = 3E \]

### Final Answer:
\[ \Delta E = 3E \]