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

Explanation

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

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the electron,
- \( E \) is the kinetic energy of the electron.

2. **Relate Wavelength to Kinetic Energy**:
From the formula, we can see that the wavelength is inversely proportional to the square root of the kinetic energy:
\[ \lambda \propto \frac{1}{\sqrt{E}} \]

3. **Define Initial Kinetic Energy**:
Let the initial kinetic energy be \( E \). Therefore, the initial wavelength (\( \lambda \)) can be expressed as:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

4. **Determine New Kinetic Energy**:
If the kinetic energy doubles, the new kinetic energy (\( E' \)) is:
\[ E' = 2E \]

5. **Calculate New Wavelength**:
Substitute the new kinetic energy into the de Broglie wavelength formula:
\[ \lambda' = \frac{h}{\sqrt{2mE'}} = \frac{h}{\sqrt{2m(2E)}} = \frac{h}{\sqrt{4mE}} = \frac{h}{2\sqrt{mE}} = \frac{1}{2} \cdot \frac{h}{\sqrt{2mE}} = \frac{1}{2} \lambda \]

6. **Find the Factor of Change**:
To find how the wavelength changes, we compare the new wavelength (\( \lambda' \)) to the initial wavelength (\( \lambda \)):
\[ \frac{\lambda'}{\lambda} = \frac{1/2 \cdot \lambda}{\lambda} = \frac{1}{2} \]

### Conclusion:
The de Broglie wavelength changes by a factor of \( \frac{1}{2} \) when the kinetic energy of a free electron doubles.