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 finding the speed of an electron as it enters the p-side of a p-n junction, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Speed of the electron approaching the p-n junction from the n-side (v1) = \(5 \times 10^5 \, \text{m/s}\)
- Mass of the electron (m) = \(9.1 \times 10^{-31} \, \text{kg}\)
- Charge of the electron (e) = \(1.6 \times 10^{-19} \, \text{C}\)
- Potential difference across the junction (V) = \(0.5 \, \text{V}\) (assumed for this example)

### Step 2: Write the energy conservation equation
When the electron moves from the n-side to the p-side, its kinetic energy and potential energy change. The conservation of energy can be expressed as:
\[ \frac{1}{2} m v_1^2 = \frac{1}{2} m v_2^2 + eV \]
Where:
- \(v_1\) is the initial speed (before entering the p-side)
- \(v_2\) is the final speed (after entering the p-side)
- \(eV\) is the potential energy gained by the electron when it crosses the junction.

### Step 3: Rearrange the equation to solve for \(v_2\)
Rearranging the equation gives:
\[ \frac{1}{2} m v_2^2 = \frac{1}{2} m v_1^2 - eV \]
Multiplying through by 2 to eliminate the fraction:
\[ m v_2^2 = m v_1^2 - 2eV \]
Now, divide by \(m\):
\[ v_2^2 = v_1^2 - \frac{2eV}{m} \]

### Step 4: Substitute the known values
Substituting the known 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 \(v_1^2\):
\[ (5 \times 10^5)^2 = 2.5 \times 10^{11} \, \text{m}^2/\text{s}^2 \]

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

Now substituting back:
\[ v_2^2 = 2.5 \times 10^{11} - 1.76 \times 10^{11} = 0.74 \times 10^{11} \]

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

### Final Answer
The speed of the electron as it enters the p-side is approximately \(8.6 \times 10^5 \, \text{m/s}\).

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