By what factor does the de-Broglie wavelength of a free electron changes if its kinetic energy is doubled ?

Options

A. `(1)/(2)`
B. 2
C. `(1)/(sqrt(2))`
D. `sqrt(2)`

Answer

To solve the problem of how the de-Broglie wavelength of a free electron changes when its kinetic energy is doubled, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and momentum
The kinetic energy (K.E.) of a free electron is given by the equation:
\[ K.E. = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the electron and \(v\) is its velocity.

### Step 2: Express momentum in terms of kinetic energy
The momentum \(p\) of the electron can be expressed as:
\[ p = mv \]
From the kinetic energy equation, we can derive the momentum in terms of kinetic energy:
\[ K.E. = \frac{1}{2} mv^2 \implies v = \sqrt{\frac{2K.E.}{m}} \]
Thus, substituting for \(v\) in the momentum equation:
\[ p = m \sqrt{\frac{2K.E.}{m}} = \sqrt{2m \cdot K.E.} \]

### Step 3: Write the de-Broglie wavelength formula
The de-Broglie wavelength \(\lambda\) is given by:
\[ \lambda = \frac{h}{p} \]
Substituting for \(p\):
\[ \lambda = \frac{h}{\sqrt{2m \cdot K.E.}} \]

### Step 4: Analyze the effect of doubling the kinetic energy
If the kinetic energy is doubled, we have:
\[ K.E. \rightarrow K.E.' = 2K.E. \]
Substituting this into the de-Broglie wavelength equation:
\[ \lambda' = \frac{h}{\sqrt{2m \cdot (2K.E.)}} = \frac{h}{\sqrt{4m \cdot K.E.}} = \frac{h}{2\sqrt{m \cdot K.E.}} = \frac{1}{2} \cdot \frac{h}{\sqrt{2m \cdot K.E.}} = \frac{1}{2} \lambda \]

### Step 5: Conclusion
The new de-Broglie wavelength \(\lambda'\) is half of the original wavelength \(\lambda\). Therefore, the factor by which the de-Broglie wavelength changes when the kinetic energy is doubled is:
\[ \text{Factor} = \frac{\lambda'}{\lambda} = \frac{1}{2} \]

Thus, the de-Broglie wavelength of a free electron decreases by a factor of 2 when its kinetic energy is doubled.