In a p-type semiconductor,the acceptor level | vedclass

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(D) Given energy required to produce a hole,$E = 50 	ext{ meV} = 50 	imes 10^{-3} 	imes 1.6 	imes 10^{-19} 	ext{ J} = 8.0 	imes 10^{-21} 	ext{

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$24.8$

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(D) Given energy required to produce a hole,$E = 50 	ext{ meV} = 50 	imes 10^{-3} 	imes 1.6 	imes 10^{-19} 	ext{ J} = 8.0 	imes 10^{-21} 	ext{ J}$.Planck's constant,$h = 6.6 	imes 10^{-34} 	ext{ Js}$.Speed of light,$c = 3 	imes 10^8 	ext{ m/s}$.The energy of a photon is given by $E = \frac{hc}{\lambda}$.To find the maximum wavelength $\lambda$,we use $\lambda = \frac{hc}{E}$.Substituting the values:$\lambda = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{8.0 	imes 10^{-21}}$$\lambda = \frac{19.8 	imes 10^{-26}}{8.0 	imes 10^{-21}}$$\lambda = 2.475 	imes 10^{-5} 	ext{ m} = 24.75 	imes 10^{-6} 	ext{ m} = 24.75 \mu 	ext{m}$.Rounding to the nearest option,the maximum wavelength is $24.8 \mu 	ext{m}$.

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