The band gap in a semiconductor is 0.6 eV. Th | vedclass

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(C) The energy of a photon required to create an electron-hole pair in a semiconductor must be at least equal to the band gap energy $(E_g)$.$E_g = 0.

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$2070$

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(C) The energy of a photon required to create an electron-hole pair in a semiconductor must be at least equal to the band gap energy $(E_g)$.$E_g = 0.6 eV$We know that the energy of a photon is given by $E = \frac{hc}{\lambda}$.To find the maximum wavelength $(\lambda_{max})$, we set the photon energy equal to the band gap energy:$E_g = \frac{hc}{\lambda_{max}}$$\lambda_{max} = \frac{hc}{E_g}$Given $hc = 1242 eV-nm$ and $E_g = 0.6 eV$:$\lambda_{max} = \frac{1242}{0.6} nm = 2070 nm$If the wavelength is greater than $2070 nm$, the photon energy will be less than $0.6 eV$, which is insufficient to excite an electron from the valence band to the conduction band. Therefore, $2070 nm$ is the maximum wavelength.

The band gap in a semiconductor is $0.6 eV$. The maximum wavelength of electromagnetic radiation which can create a hole-electron pair in this semiconductor is equal to [Use $hc = 1242 eV-nm$]. (in $nm$)

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