Question:
An electron in n-region of a p-n junction moves towards the junction with a speed of \(5 \times 10^5\ \text{ms}^{-1}\). If the barrier potential of the junction is \(0.45\ \text{V}\), then the speed with which the electron enters the p-region after penetration through the barrier is
(Charge of the electron \(= 1.6 \times 10^{-19}\ \text{C}\) and mass of the electron \(= 9 \times 10^{-31}\ \text{kg}\))
An electron in n-region of a p-n junction moves towards the junction with a speed of \(5 \times 10^5\ \text{ms}^{-1}\). If the barrier potential of the junction is \(0.45\ \text{V}\), then the speed with which the electron enters the p-region after penetration through the barrier is
(Charge of the electron \(= 1.6 \times 10^{-19}\ \text{C}\) and mass of the electron \(= 9 \times 10^{-31}\ \text{kg}\))
(Charge of the electron \(= 1.6 \times 10^{-19}\ \text{C}\) and mass of the electron \(= 9 \times 10^{-31}\ \text{kg}\))
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Remember that majority carriers (electrons in n-type) must climb an energy hill to cross the junction. Therefore, their kinetic energy decreases.
Updated On: Mar 26, 2026
- \(3 \times 10^5\ \text{ms}^{-1}\)
- \(5 \times 10^5\ \text{ms}^{-1}\)
- \(4 \times 10^5\ \text{ms}^{-1}\)
- \(6 \times 10^5\ \text{ms}^{-1}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
When an electron moves from the n-region to the p-region across the depletion layer, it has to overcome the potential barrier. The electric field in the depletion region opposes the motion of majority charge carriers (electrons from n-side). Consequently, the electron loses kinetic energy equal to the potential energy barrier.
Step 2: Apply Law of Conservation of Energy:
\[ \text{Initial K.E.} - \text{Work done against barrier} = \text{Final K.E.} \] \[ \frac{1}{2} m u^2 - e V_0 = \frac{1}{2} m v^2 \] Where: \(u\) is the initial speed \(= 5 \times 10^5\ \text{m/s}\) \(v\) is the final speed \(V_0\) is the barrier potential \(= 0.45\ \text{V}\)
Step 3: Solve for Final Speed (\(v\)):
Rearranging the equation: \[ v^2 = u^2 - \frac{2eV_0}{m} \] Substitute the values: \[ u^2 = (5 \times 10^5)^2 = 25 \times 10^{10} \] \[ \frac{2eV_0}{m} = \frac{2 \times (1.6 \times 10^{-19}) \times 0.45}{9 \times 10^{-31}} \] \[ \frac{2eV_0}{m} = \frac{1.44 \times 10^{-19}}{9 \times 10^{-31}} = 0.16 \times 10^{12} = 16 \times 10^{10} \] Now, calculate \(v^2\): \[ v^2 = 25 \times 10^{10} - 16 \times 10^{10} \] \[ v^2 = 9 \times 10^{10} \] \[ v = \sqrt{9 \times 10^{10}} = 3 \times 10^5\ \text{m/s} \] Final Answer:
The speed is \(3 \times 10^5\ \text{ms}^{-1}\).
When an electron moves from the n-region to the p-region across the depletion layer, it has to overcome the potential barrier. The electric field in the depletion region opposes the motion of majority charge carriers (electrons from n-side). Consequently, the electron loses kinetic energy equal to the potential energy barrier.
Step 2: Apply Law of Conservation of Energy:
\[ \text{Initial K.E.} - \text{Work done against barrier} = \text{Final K.E.} \] \[ \frac{1}{2} m u^2 - e V_0 = \frac{1}{2} m v^2 \] Where: \(u\) is the initial speed \(= 5 \times 10^5\ \text{m/s}\) \(v\) is the final speed \(V_0\) is the barrier potential \(= 0.45\ \text{V}\)
Step 3: Solve for Final Speed (\(v\)):
Rearranging the equation: \[ v^2 = u^2 - \frac{2eV_0}{m} \] Substitute the values: \[ u^2 = (5 \times 10^5)^2 = 25 \times 10^{10} \] \[ \frac{2eV_0}{m} = \frac{2 \times (1.6 \times 10^{-19}) \times 0.45}{9 \times 10^{-31}} \] \[ \frac{2eV_0}{m} = \frac{1.44 \times 10^{-19}}{9 \times 10^{-31}} = 0.16 \times 10^{12} = 16 \times 10^{10} \] Now, calculate \(v^2\): \[ v^2 = 25 \times 10^{10} - 16 \times 10^{10} \] \[ v^2 = 9 \times 10^{10} \] \[ v = \sqrt{9 \times 10^{10}} = 3 \times 10^5\ \text{m/s} \] Final Answer:
The speed is \(3 \times 10^5\ \text{ms}^{-1}\).
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