Neglecting veriation of mass with velocity , the wavelength associated with an electron having a kinetic energy E is proportional to

Options

A. ` E^(1//2)`
B. E
C. ` E^(-1//2)`
D. `E^(-1//2)`

Answer

To solve the problem of determining how the wavelength associated with an electron having kinetic energy \( E \) is proportional, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength**:
The de Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Express Momentum in Terms of Mass and Velocity**:
The momentum \( p \) of an electron can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

3. **Relate Kinetic Energy to Velocity**:
The kinetic energy \( E \) of the electron is given by:
\[ E = \frac{1}{2} mv^2 \]
Rearranging this equation gives us:
\[ v^2 = \frac{2E}{m} \]
Taking the square root, we find:
\[ v = \sqrt{\frac{2E}{m}} \]

4. **Substitute Velocity Back into Momentum**:
Now, substituting \( v \) back into the momentum equation:
\[ p = m \cdot \sqrt{\frac{2E}{m}} = \sqrt{2mE} \]

5. **Substitute Momentum into the Wavelength Formula**:
Now substituting \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

6. **Identify the Proportional Relationship**:
From the equation \( \lambda = \frac{h}{\sqrt{2mE}} \), we can see that:
\[ \lambda \propto \frac{1}{\sqrt{E}} \]
This indicates that the wavelength \( \lambda \) is inversely proportional to the square root of the kinetic energy \( E \).

### Conclusion:
Thus, the wavelength associated with an electron having kinetic energy \( E \) is proportional to \( \frac{1}{\sqrt{E}} \).