If an electron approaches the p-n junction from the n-side with a speed of `5 xx 10^5 m s^(-1)` ,with what speed will it enter the p-side?

Options

A. `5xx10^5` m/s
B. `2.5xx10^6` m/s
C. `2.7xx10^5` m/s
D. `1xx10^5` m/s

Explanation

To solve the problem of an electron approaching a p-n junction from the n-side with a speed of \(5 \times 10^5 \, \text{m/s}\) and determining its speed upon entering the p-side, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the System
We have an electron moving towards a p-n junction. The electron has an initial speed \(v_1 = 5 \times 10^5 \, \text{m/s}\) as it approaches the junction from the n-side. When it enters the p-side, it will experience a change in potential energy due to the electric field present in the depletion region.

### Step 2: Apply Conservation of Energy
The total mechanical energy of the electron must be conserved. The initial kinetic energy of the electron will be converted into kinetic energy and potential energy when it enters the p-side.

The equation can be expressed as:
\[ \frac{1}{2} m v_1^2 = \frac{1}{2} m v_2^2 + qV \]
where:
- \(m\) is the mass of the electron,
- \(v_1\) is the initial speed,
- \(v_2\) is the speed on the p-side,
- \(q\) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)),
- \(V\) is the potential difference across the junction.

### Step 3: Rearrange the Equation
Rearranging the equation to solve for \(v_2\):
\[ \frac{1}{2} m v_2^2 = \frac{1}{2} m v_1^2 - qV \]
\[ v_2^2 = v_1^2 - \frac{2qV}{m} \]

### Step 4: Substitute Known Values
Given:
- \(v_1 = 5 \times 10^5 \, \text{m/s}\)
- \(m = 9.1 \times 10^{-31} \, \text{kg}\)
- \(q = 1.6 \times 10^{-19} \, \text{C}\)
- Assume \(V = 0.5 \, \text{V}\) (as mentioned in the transcript).

Now substituting these values into the equation:
\[ v_2^2 = (5 \times 10^5)^2 - \frac{2 \times (1.6 \times 10^{-19}) \times (0.5)}{9.1 \times 10^{-31}} \]

Calculating \(v_1^2\):
\[ (5 \times 10^5)^2 = 2.5 \times 10^{11} \]

Calculating the potential energy term:
\[ \frac{2 \times (1.6 \times 10^{-19}) \times (0.5)}{9.1 \times 10^{-31}} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 1.76 \times 10^{11} \]

### Step 5: Calculate \(v_2^2\)
Now substituting back:
\[ v_2^2 = 2.5 \times 10^{11} - 1.76 \times 10^{11} = 0.74 \times 10^{11} \]

### Step 6: Solve for \(v_2\)
Taking the square root:
\[ v_2 = \sqrt{0.74 \times 10^{11}} \approx 8.6 \times 10^5 \, \text{m/s} \]

### Final Answer
Thus, the speed of the electron as it enters the p-side is approximately:
\[ v_2 \approx 2.7 \times 10^5 \, \text{m/s} \]