The de - Broglie wavelength `lambda` associated with an electron having kinetic energy `E` is given by the expression

Options

A. `(h)/(sqrt( 2 m E))`
B. `( 2h)/(m E)`
C. `2 mhE`
D. `( 2 sqrt(2 mE))/(h)`

Explanation

To find the de Broglie wavelength (λ) associated with an electron having kinetic energy (E), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between kinetic energy and momentum**:
The kinetic energy (E) of an electron can be expressed as:
\[ E = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the electron and \(v\) is its velocity.

2. **Express momentum in terms of kinetic energy**:
We know that momentum \(p\) is given by:
\[ p = mv \]
We can rearrange the kinetic energy equation to solve for \(v\):
\[ v = \sqrt{\frac{2E}{m}} \]
Substituting this expression for \(v\) into the momentum equation gives us:
\[ p = m \cdot \sqrt{\frac{2E}{m}} = \sqrt{2mE} \]

3. **Use the de Broglie wavelength formula**:
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant.

4. **Substitute the expression for momentum into the de Broglie wavelength formula**:
Now, substituting \(p = \sqrt{2mE}\) into the de Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

5. **Final expression**:
Therefore, the de Broglie wavelength associated with an electron having kinetic energy \(E\) is:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

### Summary:
The de Broglie wavelength associated with an electron having kinetic energy \(E\) is given by:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]