The initial kinetic energy and wavelength for a free electron are E and `lamda` respectively. If E is made 4E for the particle then its wavelength will become _______ `lamda`

Explanation

To solve the problem, we need to understand the relationship between the kinetic energy of a free electron and its wavelength. The wavelength of a particle is given by the de Broglie wavelength formula:

\[ \lambda = \frac{h}{\sqrt{2mE}} \]

where:
- \( \lambda \) is the wavelength,
- \( h \) is Planck's constant,
- \( m \) is the mass of the electron,
- \( E \) is the kinetic energy of the electron.

### Step-by-Step Solution:

1. **Initial Wavelength**: We start with the initial kinetic energy \( E \) and the corresponding wavelength \( \lambda \):
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

2. **Final Kinetic Energy**: The problem states that the kinetic energy is increased to \( 4E \). We need to find the new wavelength \( \lambda' \) corresponding to this new energy:
\[ \lambda' = \frac{h}{\sqrt{2m(4E)}} \]

3. **Simplifying the Final Wavelength**: We can simplify the expression for \( \lambda' \):
\[ \lambda' = \frac{h}{\sqrt{8mE}} = \frac{h}{\sqrt{4 \cdot 2mE}} = \frac{h}{2\sqrt{2mE}} = \frac{1}{2} \cdot \frac{h}{\sqrt{2mE}} = \frac{1}{2} \lambda \]

4. **Final Result**: Thus, the final wavelength \( \lambda' \) is:
\[ \lambda' = \frac{\lambda}{2} \]

### Conclusion:
The new wavelength when the kinetic energy is increased to \( 4E \) will be \( \frac{\lambda}{2} \).