JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
\(LED\) is constructed from \(Ga-As-P\) semiconducting material. The energy gap of this \(LED\) is \(1.9\, eV\). Calculate the wavelength of light emitted and its colour. \(\left[ h =6.63 \times 10^{-34} \;Js \right.\) and \(\left. c =3 \times 10^{8}\; ms ^{-1}\right]\)
- A \(1046\, nm\) and red colour
- B \(654\, nm\) and orange colour
- C \(1046\, nm\) and blue colour
- D \(654\, nm\) and red colour
Answer & Solution
Correct Answer
(D) \(654\, nm\) and red colour
Step-by-step Solution
Detailed explanation
\(\lambda=\frac{ hc }{ E }=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{1.9 \times 1.6 \times 10^{-19}}=6.54 \times 10^{-7}\) \(=654 nm\) Red color
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