The wavelength `lambda` of de Broglie waves associated with an electron (mass `m`, charge) accelerated through a potential difference of `V` is given by (`h` is plank's constant):

Options

A. `lambda=h//mV`
B. `lambda=h//2 me V`
C. `lambda=h//sqrt(meV)`
D. `lambda=h//sqrt(2meV)`

Answer

To find the de Broglie wavelength (λ) of an electron that has been accelerated through a potential difference (V), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Kinetic Energy**:
When an electron is accelerated through a potential difference (V), it gains kinetic energy. The kinetic energy (KE) gained by the electron can be expressed as:
\[ KE = eV \]
where \( e \) is the charge of the electron.

2. **Relate Kinetic Energy to Momentum**:
The momentum (P) of an object can be related to its kinetic energy using the formula:
\[ KE = \frac{P^2}{2m} \]
Rearranging this gives us:
\[ P = \sqrt{2m \cdot KE} \]

3. **Substituting Kinetic Energy**:
Now, substitute the expression for kinetic energy (KE = eV) into the momentum equation:
\[ P = \sqrt{2m \cdot eV} \]

4. **Using 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.

5. **Substituting Momentum into the Wavelength Formula**:
Now substitute the expression for momentum (P) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot eV}} \]

6. **Final Expression**:
Thus, the wavelength of the de Broglie waves associated with an electron accelerated through a potential difference \( V \) is:
\[ \lambda = \frac{h}{\sqrt{2meV}} \]

### Conclusion:
The correct expression for the de Broglie wavelength of an electron accelerated through a potential difference \( V \) is \( \lambda = \frac{h}{\sqrt{2meV}} \).