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JEE Mains · Physics · STD 12 - 14. Semicondutor electronics

बैंगनी (तरंगदैर्ध्य \(=4000 \mathring A)\) \(LED\) बनाने के लिए, अर्द्धचालक पदार्थ का ऊर्जा बैंड अंतराल \(............eV\) होगा। (उत्तर निकटतम पूर्णाक में दो).

  1. A \(3\)
  2. B \(2\)
  3. C \(4\)
  4. D \(1\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(3\)

Step-by-step Solution

Detailed explanation

\(E _{ g }=\frac{ hc }{\lambda}=\frac{1242}{\lambda( nm )}=\frac{1242}{400}=3.105\) Answer rounded to \(3\,eV\)
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