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(A) The bandgap energy $E_g$ corresponds to the energy of the incident photon with the maximum wavelength $\lambda = 2480 \, nm$ that can excite an el

Vedclass · Accepted Answer

$0.5$

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(A) The bandgap energy $E_g$ corresponds to the energy of the incident photon with the maximum wavelength $\lambda = 2480 \, nm$ that can excite an electron across the bandgap.Using the formula $E = \frac{hc}{\lambda}$,where $h = 6.63 	imes 10^{-34} \, J \cdot s$,$c = 3 	imes 10^8 \, m/s$,and $1 \, eV = 1.6 	imes 10^{-19} \, J$:$E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{2480 	imes 10^{-9} 	imes 1.6 	imes 10^{-19}} \, eV$$E = \frac{19.89 	imes 10^{-26}}{3968 	imes 10^{-28}} \, eV$$E = \frac{1989}{3968} \approx 0.5 \, eV$.Thus,the bandgap of the semiconductor is $0.5 \, eV$.

When radiation with a wavelength shorter than $2480 \, nm$ is incident on a semiconductor,its conductivity increases. The bandgap of this semiconductor is ....... $eV$.

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