The energy gap in silicon is 1.1 eV, whereas in diamond it is 6eV

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Prince Budha · Accepted Answer

Solution : The energy gap in Silicon (left ( E_{g} right )_{Si}) = 1.1 eV = 1.1 (times) 1.6 (times) 10-19 j Energy gap in diamond (left ( E_{g} right )_{dia} ) = 6 eV = 6 (times )1.6( times ) 10-19j Wavelength of visible light = 4000 A° to 7000 A° Wee know that, Band gap energy (Eg) = (frac{hc}{lambda_{c}}) For Silicon (left ( E_{g} right )_{Si}) = (frac{hc}{left (lambda_{c} right )}_{Si}) ( left (lambda right )_{Si}) = (frac {6.626 times 10^{-34} times 3 times 10^{8}}{1.1 times 1.6 times 10^{-19}}) ( left (lambda right )_{Si}) = 1.1290( times 10^{-10} )= 1129 nm ( therefore left (lambda right )_{Si}) = 11290 A° Hence all visible lights are absorbed since ( left (lambda_{c} right )_{Si}) < ( lambda_{visible} ) but it can transmit infrared light having wavelength ≈ 1.1 ( times ) 10-6 m. For diamond, ( left (lambda_{c} right )_{dia}) = ( frac {hc}{left ( E_{g} right )_{dia}}) ( left (lambda_{c} right )_{dia}) = (frac{6.626 times 10^{-34} times 3 times 10^{8}}{6 times 1.6 times 10^{-19}} ) ( left (lambda_{c} right )_{dia}) = 2.066 (times ) 10-7 m = 2066 (times ) 10-10 m ( therefore left (lambda_{c} right )_{dia}) = 2066 A° = 206 nm ( therefore left (lambda_{c} right )_{dia}) < ( lambda_{visible}) i.e Ephoton (visible light) <Eg Hence, in the case of diamond all visible lights are transmitted by such type of non metallic materials. Thus, Diamond appears transparent and colorless.

The energy gap in silicon is 1.1 eV, whereas in diamond it is 6eV. What conclusion can you draw about the transparency of the two materials to visible light(4000 A° to 7000 A°)?

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The energy gap in silicon is 1.1 eV, whereas in diamond it is 6eV. What conclusion can you draw about the transparency of the two materials to visible light(4000 A° to 7000 A°)?

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