Why cannot we use Si and Ge in fabrication of visible LEDs ?

The energy gap for Si is 1 . 1 eV and for Ge is 0 . 7 eV . Both of the semiconductor are having energy gap less than 1 . 8 eV . Thus, instead of producing photon of visible wavelength, they produce heat in the form of Infrared rays. Hence, Si and Ge are not used in fabrication of visible LEDs.

Three photo diodes D 1 , D 2 and D 3 are made of semiconductors having band gaps of 2 . 5 eV , 2 eV and 3 eV , respectively. Which ones will be able to detect light of wavelength 600 n m ? Given Planck constant is h = 6 . 62 × 10 - 34 m 2 k g s - 1 ,speed of light c = 3 × 10 8 m s - 1

Given that: λ = 600 n m = 600 × 10 - 9 m E = h c λ = 6 . 62 × 10 - 34 × 3 × 10 8 600 × 10 - 9 × 1 . 6 × 10 - 19 eV = 6 . 62 × 3 × 10 - 34 + 8 6 × 1 . 6 × 10 - 26 eV = 2 . 06 eV The incident radiation is detected by photodiode D 2 has band gap or knee voltage 2 eV only which is less than incident radiation of 2 . 06 eV . The photo diode is able to detect the light of given wavelength, if the energy of incident radiation is more than the band gap of the photo diode. Therefore, D 1 and D 3 are not able to detect incident light.