A potential barrier V volts exists across a P-N junction. The thickness of the depletion region is 'd'. An electron with velocity 'v' approaches P-N junction from N-side. The velocity of the electron across the junction is

Options

A. `sqrt(v^2 +(2Ve)/(m))`
B. `sqrt((v^2 -( 2Ve )/( m ))`
C. `v`
D. `sqrt((2Ve)/( m))`

Answer

To solve the problem of finding the velocity of an electron as it crosses a P-N junction with a potential barrier \( V \) volts and a depletion region thickness \( d \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let \( v \) be the initial velocity of the electron approaching the P-N junction.
- Let \( V \) be the potential barrier across the junction.
- Let \( d \) be the thickness of the depletion region.
- Let \( m \) be the mass of the electron.
- Let \( e \) be the charge of the electron.

2. **Apply the Kinematic Equation:**
- We use the kinematic equation:
\[ v_f^2 = v^2 + 2a s \]
- Here, \( v_f \) is the final velocity of the electron after crossing the junction, \( a \) is the acceleration, and \( s \) is the displacement (which is \( d \)).

3. **Determine the Acceleration:**
- The force acting on the electron due to the electric field \( E \) in the depletion region is given by:
\[ F = eE \]
- The electric field \( E \) can be expressed as:
\[ E = \frac{V}{d} \]
- Therefore, the acceleration \( a \) of the electron can be calculated as:
\[ a = \frac{F}{m} = \frac{eE}{m} = \frac{eV}{md} \]

4. **Substitute Acceleration into the Kinematic Equation:**
- Substitute \( a \) into the kinematic equation:
\[ v_f^2 = v^2 + 2\left(\frac{eV}{md}\right)d \]
- This simplifies to:
\[ v_f^2 = v^2 + \frac{2eV}{m} \]

5. **Solve for Final Velocity:**
- Rearranging gives:
\[ v_f^2 = v^2 - \frac{2eV}{m} \]
- Taking the square root gives us the final velocity:
\[ v_f = \sqrt{v^2 - \frac{2eV}{m}} \]

6. **Conclusion:**
- The final velocity of the electron as it crosses the junction is:
\[ v_f = \sqrt{v^2 - \frac{2eV}{m}} \]
- Thus, the correct option is \( \sqrt{v^2 - \frac{2eV}{m}} \).