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Explain why elemental semiconductor cannot be used to make visible LEDs.

 

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In an elemental semiconductor, the bandgap is such that the emission is in the infrared region and not in the visible region.

\lambda=\frac{hc}{E_{g}}=\frac{1242eVnm}{E_{g}}

for Si; E_{g}=1.1eV, \lambda=\frac{1242}{1.1}=1129nm

for Ge; E_{g}=0.7eV, \lambda=\frac{1242}{0.7}=1725nm

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