TS EAMCET · Physics · Semiconductors
An electron in n-region of a p-n junction moves towards the junction with a speed of \(5 \times 10^5 \mathrm{~ms}^{-1}\). If the barrier potential of the junction is 0.45 V, then the speed with which the electron enters the p-region after penetration through the barrier is (Charge of the electron \(=1.6 \times 10^{-19} \mathrm{C}\) and mass of the electron \(=9 \times 10^{-31} \mathrm{~kg}\) )
- A \(3 \times 10^5 \mathrm{~ms}^{-1}\)
- B \(5 \times 10^5 \mathrm{~ms}^{-1}\)
- C \(4 \times 10^5 \mathrm{~ms}^{-1}\)
- D \(6 \times 10^5 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(A) \(3 \times 10^5 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
\( \frac{1}{2} m v_1^2 - e V_b = \frac{1}{2} m v_2^2 \) \( \frac{1}{2} (9 \times 10^{-31}) (5 \times 10^5)^2 - (1.6 \times 10^{-19}) (0.45) = \frac{1}{2} (9 \times 10^{-31}) v_2^2 \) \( 1.125 \times 10^{-19} - 0.72 \times 10^{-19} = 4.5 \times 10^{-31} v_2^2 \)…
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