JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
If the potential barrier across a \(p - n\) junction is \(0.6\,V\). Then the electric field intensity, in the depletion region having the width of \(6 \times 10^{-6} m\), will be \(......\times 10^{5} N / C\).
- A \(0\)
- B \(1\)
- C \(10\)
- D \(100\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(E =\frac{ V }{ d }=\frac{\text { Potential barrier Across Junction }}{\text { width of Depletion layer }}\) \(=\frac{0.6\,V }{6 \times 10^{-6}\,m }=1 \times 10^{5}\,V / m\) \(=1 \times 10^{5}\,N / C\)
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