A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If c and m are charge and mass of an electron repectively, then the value of `h//lambda` (where `lambda` is wavelength associated with electron wave) is given by :

Options

A. 2meV
B. `sqrt(me V)`
C. `sqrt(2me V)`
D. meV

Answer

To solve the problem, we need to find the value of \( \frac{h}{\lambda} \), where \( \lambda \) is the wavelength associated with the electron wave. We will use the relationship between kinetic energy, potential difference, and the de Broglie wavelength.

### Step-by-Step Solution

1. **Understand the Kinetic Energy of the Electron**:
The kinetic energy (KE) of an electron accelerated through a potential difference \( V \) is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron.

2. **Relate Kinetic Energy to Velocity**:
The kinetic energy can also be expressed in terms of the mass \( m \) of the electron and its velocity \( u \):
\[ KE = \frac{1}{2} mu^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ eV = \frac{1}{2} mu^2 \]

3. **Solve for Velocity**:
Rearranging the equation to solve for \( u^2 \):
\[ mu^2 = 2eV \]
\[ u^2 = \frac{2eV}{m} \]
Taking the square root to find \( u \):
\[ u = \sqrt{\frac{2eV}{m}} \]

4. **Use the de Broglie Wavelength Formula**:
The de Broglie wavelength \( \lambda \) of a particle is given by:
\[ \lambda = \frac{h}{mv} \]
Substituting \( u \) for \( v \):
\[ \lambda = \frac{h}{mu} \]

5. **Substitute for Velocity**:
Substitute \( u = \sqrt{\frac{2eV}{m}} \) into the de Broglie wavelength equation:
\[ \lambda = \frac{h}{m\sqrt{\frac{2eV}{m}}} \]
Simplifying this gives:
\[ \lambda = \frac{h}{\sqrt{2emV}} \]

6. **Find \( \frac{h}{\lambda} \)**:
Taking the reciprocal to find \( \frac{h}{\lambda} \):
\[ \frac{h}{\lambda} = \sqrt{2emV} \]

### Final Result
Thus, the value of \( \frac{h}{\lambda} \) is:
\[ \frac{h}{\lambda} = \sqrt{2emV} \]