A potential barrier of 0.3 V exists across a p-n junction. An electron with speed `5 xx 10^5` m/s approaches this p-n junction from n-side, what will be its speed on entering the p-side?

Options

A. `3 xx 10^5 m//s`
B. `3.8 xx 10^5 m//s`
C. `5xx 10^5 m//s`
D. `2.6 xx 10^3 m//s`

Explanation

To solve the problem, we need to determine the speed of an electron as it crosses a potential barrier of 0.3 V at a p-n junction. The initial speed of the electron is given as \(5 \times 10^5 \, \text{m/s}\).

### Step-by-Step Solution:

1. **Understanding the Potential Barrier**:
The potential barrier of 0.3 V means that the electron will lose some kinetic energy as it crosses this barrier. The energy lost can be calculated using the formula:
\[ \text{Energy lost} = e \cdot V \]
where \(e\) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)) and \(V\) is the potential barrier (0.3 V).

2. **Calculating the Energy Lost**:
\[ \text{Energy lost} = (1.6 \times 10^{-19} \, \text{C}) \cdot (0.3 \, \text{V}) = 4.8 \times 10^{-20} \, \text{J} \]

3. **Calculating the Initial Kinetic Energy**:
The initial kinetic energy (\(KE_i\)) of the electron can be calculated using the formula:
\[ KE_i = \frac{1}{2} m v^2 \]
where \(m\) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)) and \(v\) is the initial speed (\(5 \times 10^5 \, \text{m/s}\)).
\[ KE_i = \frac{1}{2} (9.11 \times 10^{-31} \, \text{kg}) (5 \times 10^5 \, \text{m/s})^2 \]
\[ KE_i = \frac{1}{2} (9.11 \times 10^{-31}) (2.5 \times 10^{11}) = 1.14 \times 10^{-19} \, \text{J} \]

4. **Calculating the Final Kinetic Energy**:
The final kinetic energy (\(KE_f\)) after crossing the barrier will be:
\[ KE_f = KE_i - \text{Energy lost} \]
\[ KE_f = (1.14 \times 10^{-19} \, \text{J}) - (4.8 \times 10^{-20} \, \text{J}) = 6.56 \times 10^{-20} \, \text{J} \]

5. **Finding the Final Speed**:
Now we can find the final speed (\(v_f\)) using the final kinetic energy:
\[ KE_f = \frac{1}{2} m v_f^2 \]
Rearranging gives:
\[ v_f = \sqrt{\frac{2 \cdot KE_f}{m}} \]
Substituting the values:
\[ v_f = \sqrt{\frac{2 \cdot (6.56 \times 10^{-20})}{9.11 \times 10^{-31}}} \]
\[ v_f = \sqrt{\frac{1.312 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx \sqrt{1.44 \times 10^{11}} \approx 1.2 \times 10^5 \, \text{m/s} \]

### Final Answer:
The speed of the electron on entering the p-side is approximately \(1.2 \times 10^5 \, \text{m/s}\).