If the kinetic energy of a free electron doubles , its de - Broglie wavelength changes by the factor

Options

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

Explanation

To solve the problem, we need to analyze how the de Broglie wavelength of a free electron changes when its kinetic energy doubles. Here is the step-by-step solution:

### Step 1: Understand the relationship between kinetic energy and momentum
The kinetic energy (KE) of a free electron is given by the formula:
\[ KE = \frac{p^2}{2m} \]
where \( p \) is the momentum and \( m \) is the mass of the electron.

### Step 2: Express momentum in terms of kinetic energy
From the kinetic energy formula, we can express momentum as:
\[ p = \sqrt{2m \cdot KE} \]

### Step 3: Determine the initial and final kinetic energy
Let the initial kinetic energy be \( KE_1 = k \) and the final kinetic energy when it doubles be \( KE_2 = 2k \).

### Step 4: Calculate the initial and final momentum
Using the expression for momentum:
- Initial momentum:
\[ p_1 = \sqrt{2m \cdot k} \]
- Final momentum:
\[ p_2 = \sqrt{2m \cdot (2k)} = \sqrt{4mk} = 2\sqrt{mk} \]

### Step 5: Write the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant.

### Step 6: Calculate the initial and final de Broglie wavelengths
- Initial de Broglie wavelength:
\[ \lambda_1 = \frac{h}{p_1} = \frac{h}{\sqrt{2mk}} \]
- Final de Broglie wavelength:
\[ \lambda_2 = \frac{h}{p_2} = \frac{h}{2\sqrt{mk}} \]

### Step 7: Find the ratio of the wavelengths
To find how the wavelength changes, we take the ratio:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\frac{h}{2\sqrt{mk}}}{\frac{h}{\sqrt{2mk}}} = \frac{\sqrt{2mk}}{2\sqrt{mk}} = \frac{1}{\sqrt{2}} \]

### Step 8: Conclusion
Thus, the de Broglie wavelength changes by a factor of:
\[ \lambda_2 = \frac{\lambda_1}{\sqrt{2}} \]

### Final Answer
The de Broglie wavelength changes by a factor of \( \frac{1}{\sqrt{2}} \).

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