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(C) The band gap energy $E_g$ of a semiconductor is related to the maximum wavelength $\lambda$ of the photon it can detect by the formula: $E_g = \fr

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$3.1$

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(C) The band gap energy $E_g$ of a semiconductor is related to the maximum wavelength $\lambda$ of the photon it can detect by the formula: $E_g = \frac{hc}{\lambda}$.Substituting the given values:$h = 6.63 	imes 10^{-34}\, J \cdot s$$c = 3 	imes 10^{8}\, m/s$$\lambda = 400 	imes 10^{-9}\, m$$E_g = \frac{(6.63 	imes 10^{-34}) 	imes (3 	imes 10^{8})}{400 	imes 10^{-9}}\, J$$E_g = \frac{19.89 	imes 10^{-26}}{4 	imes 10^{-7}}\, J = 4.9725 	imes 10^{-19}\, J$.To convert this energy into electron-volts $(eV)$, divide by the charge of an electron $e = 1.6 	imes 10^{-19}\, C$:$E_g (in\; eV) = \frac{4.9725 	imes 10^{-19}}{1.6 	imes 10^{-19}} \approx 3.1\; eV$.

If a semiconductor photodiode can detect a photon with a maximum wavelength of $400\, nm$, then its band gap energy is (Given: Planck's constant $h = 6.63 \times 10^{-34}\, J \cdot s$, Speed of light $c = 3 \times 10^{8}\, m/s$) (in $ eV$)

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