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 \(= 9 \times 10^{-31}\ \text{kg}\))
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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.
Step 1: Understanding the Question:
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, Calculation Process \(v^2\):
\[ v^2 = 25 \times 10^{10} - 16 \times 10^{10} \]
\[ v^2 = 9 \times 10^{10} \]
\[ v = \sqrt{9 \times 10^{10}} = 3 \times 10^5\ \text{m/s} \]
Required Answer: The speed is \(3 \times 10^5\ \text{ms}^{-1}\).