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

Answer

To solve the problem, we will follow these steps:

### Step 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. The momentum \(p\) can also be expressed in terms of kinetic energy \(E\):
\[ E = \frac{p^2}{2m} \implies p = \sqrt{2mE} \]
Substituting this into the de Broglie wavelength formula, we have:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

### Step 2: Set up the equation for the new kinetic energy
When the kinetic energy is increased by \(\delta E\), the new kinetic energy \(E'\) is:
\[ E' = E + \delta E \]
The new de Broglie wavelength \(\lambda'\) is given as \(\frac{\lambda}{2}\):
\[ \lambda' = \frac{h}{\sqrt{2mE'}} = \frac{h}{\sqrt{2m(E + \delta E)}} \]
Setting \(\lambda' = \frac{\lambda}{2}\), we have:
\[ \frac{h}{\sqrt{2m(E + \delta E)}} = \frac{h}{2\sqrt{2mE}} \]

### Step 3: Simplify the equation
Cancelling \(h\) and \(2\sqrt{2m}\) from both sides gives:
\[ \frac{1}{\sqrt{E + \delta E}} = \frac{1}{2\sqrt{E}} \]
Cross-multiplying results in:
\[ 2\sqrt{E} = \sqrt{E + \delta E} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides yields:
\[ 4E = E + \delta E \]

### Step 5: Solve for \(\delta E\)
Rearranging the equation gives:
\[ \delta E = 4E - E = 3E \]

### Final Answer
Thus, the increase in kinetic energy \(\delta E\) is:
\[ \delta E = 3E \]